Trigonometry : Practice Problems

1.        If $ \displaystyle \tan \alpha + \tan \beta = a$, $ \displaystyle \cot \alpha + \cot \beta = b$ and $ \displaystyle \tan(\alpha +\beta) = c$, show that $ \displaystyle bc - ac = ab$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \alpha +\tan \beta =a\\\\\ \ \ \ \cot \alpha +\cot \beta =b\\\\\therefore \ \ \displaystyle \frac{1}{{\tan \alpha }}+\displaystyle \frac{1}{{\tan \beta }}=b\\\\\therefore \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{\tan \alpha \tan \beta }}=b\\\\\therefore \ \ \displaystyle \frac{a}{{\tan \alpha \tan \beta }}=b\\\\\therefore \ \ \tan \alpha \tan \beta =\displaystyle \frac{a}{b}\\\\\ \ \ \ \tan \left( {\alpha +\beta } \right)=c\\\\\ \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=c\\\\\therefore \ \ \displaystyle \frac{a}{{1-\displaystyle \frac{a}{b}}}=c\\\\\therefore \ \ \displaystyle \frac{{ab}}{{b-a}}=c\\\\\therefore \ \ ab=bc-ac\end{array}$

2.        If $\displaystyle \alpha +\beta =45{}^\circ $, find the value of $ \displaystyle (1+\tan \alpha )(1+\tan \beta )$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha +\beta =45{}^\circ \\\\\therefore \ \ \ \ \tan \left( {\alpha +\beta } \right)=\tan 45{}^\circ \\\\\ \ \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=1\\\\\ \ \ \ \ \tan \alpha +\tan \beta =1-\tan \alpha \tan \beta \\\\\therefore \ \ \ \tan \alpha +\tan \beta +\tan \alpha \tan \beta =1\\\\\therefore \ \ \ 1+\tan \alpha +\tan \beta +\tan \alpha \tan \beta =2\\\\\therefore \ \ \ \left( {1+\tan \alpha } \right)+\tan \beta \left( {1+\tan \alpha } \right)=2\\\\\therefore \ \ \ \left( {1+\tan \alpha } \right)\left( {1+\tan \beta } \right)=2\end{array}$

3.        If $ \displaystyle \cos \alpha +\cos \beta +\cos \gamma =0$ and $ \displaystyle \sin \alpha +\sin \beta +\sin \gamma =0$, find the value of $ \displaystyle \cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma - \alpha} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos \alpha +\cos \beta +\cos \gamma =0\\\\\ \ \ \ \sin \alpha +\sin \beta +\sin \gamma =0\\\\\therefore \ \ {{\left( {\sin \alpha +\sin \beta +\sin \gamma } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta +\cos \gamma } \right)}^{2}}=0\\\\\ \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma +2\left( {\sin \alpha \sin \beta +\sin \beta \sin \gamma +\sin \alpha \sin \gamma } \right)\ \ \ \\\ \ \ \ +{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma +2\left( {\cos \alpha \cos \beta +\cos \beta \cos \gamma +\cos \alpha \cos \gamma } \right)=0\\\\\therefore \ \ \left( {{{{\sin }}^{2}}\alpha +{{{\cos }}^{2}}\alpha +{{{\sin }}^{2}}\beta +{{{\cos }}^{2}}\beta +{{{\sin }}^{2}}\gamma {{{\cos }}^{2}}\gamma } \right)\\\ \ \ +2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta +\cos \beta \cos \gamma +\sin \beta \sin \gamma +\cos \alpha \cos \gamma +\sin \alpha \sin \gamma } \right)=0\\\\\therefore \ \ 3+2\left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)} \right]=0\\\\\therefore \ \ 2\left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)} \right]=-3\\\\\therefore \ \ \cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)=-\displaystyle \frac{3}{2}\end{array}$

4.        If $ \displaystyle \sin \alpha +\sin \beta =a$ and $ \displaystyle \cos \alpha +\cos \beta =b$, show that

(a) $ \displaystyle \sin \left( {\alpha +\beta } \right)=\frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}} \ \ $

(b) $ \displaystyle \cos \left( {\alpha +\beta } \right)=\frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \\\ \ \ \ \sin \alpha +\sin \beta =a\\\\\ \ \ \ \cos \alpha +\cos \beta =b\\\\\therefore \ \ ab=\ \left( {\sin \alpha +\sin \beta } \right)\left( {\cos \alpha +\cos \beta } \right)\\\\\ \ \ \ 2ab=2\sin \alpha \cos \alpha +2\sin \alpha \cos \beta +2\cos \alpha \sin \beta +2\sin \beta \cos \beta \\\\\ \ \ \ 2ab=2\left( {\sin \alpha \cos \beta +\cos \alpha \sin \beta } \right)+2\sin \alpha \cos \alpha +2\sin \beta \cos \beta \\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)+\sin 2\alpha +\sin 2\beta \\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)+2\sin \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\left( {\sin \alpha +\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta } \right)}^{2}}\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\sin }^{2}}\alpha +2\sin \alpha \sin \beta +{{\sin }^{2}}\beta +{{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta +2\sin \alpha \sin \beta \\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2+2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2+2\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\ \ \ \ \displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}=\displaystyle \frac{{2\sin \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}{{2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}\\\\\therefore \ \ \sin \left( {\alpha +\beta } \right)=\displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}\\\\\text{(b)}\ \\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}={{\left( {\cos \alpha +\cos \beta } \right)}^{2}}-{{\left( {\sin \alpha +\sin \beta } \right)}^{2}}\\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}={{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta -{{\sin }^{2}}\alpha -2\sin \alpha \sin \beta -{{\sin }^{2}}\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\left( {\cos \alpha \cos \beta -\sin \alpha \sin \beta } \right)+{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha +{{\cos }^{2}}\beta -{{\sin }^{2}}\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)+\cos 2\alpha +\cos 2\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)+2\cos \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\therefore \ \ \ \displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}=\displaystyle \frac{{2\cos \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}{{2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}\\\\\therefore \ \ \ \cos \left( {\alpha +\beta } \right)=\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}\end{array}$ $ \displaystyle \begin{array}{l}\underline{{\text{Alternative Method}}}\\\\\ \ \ \ \sin \alpha +\sin \beta =a\\\\\ \ \ \ 2\sin \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}=a\\\\\ \ \ \ \cos \alpha +\cos \beta =b\\\\\ \ \ \ 2\cos \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}=b\\\\\therefore \ \ \displaystyle \frac{{2\sin \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}}}{{2\cos \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}}}=\displaystyle \frac{a}{b}\\\\\therefore \ \ \tan \displaystyle \frac{{a+\beta }}{2}\\\\\ \ \ \ \text{Let}\ \text{ }a+\beta =\theta ,\ \text{then}\ \tan \displaystyle \frac{{a+\beta }}{2}=\tan \displaystyle \frac{\theta }{2}\ \\\\\ \ \ \ \text{Let}\ \tan \displaystyle \frac{\theta }{2}=t,\ \text{then}\ t=\displaystyle \frac{a}{b}\\\\\text{(a)}\sin \left( {a+\beta } \right)=\sin \theta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2t}}{{1+{{t}^{2}}}}\ \ \left[ {\text{t - formulae}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2\left( {\displaystyle \frac{a}{b}} \right)}}{{1+{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2a}}{b}}}{{\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{b}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}\\\\\text{(b)}\ \cos \left( {a+\beta } \right)=\cos \theta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{t}^{2}}}}{{1+{{t}^{2}}}}\ \ \left[ {\text{t - formulae}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}{{1+{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}}}}}{{\displaystyle \frac{{{{b}^{2}}+{{a}^{2}}}}{{{{b}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

5.        Without using tables, prove that

(a) $ \displaystyle \text{sin 47}{}^\circ +\cos 77{}^\circ =\cos 17{}^\circ $

(b) $ \displaystyle \cos 80{}^\circ +\cos 40{}^\circ -\cos 20{}^\circ =0$

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$ \displaystyle \begin{array}{l}\text{(a)}\\\ \ \ \ \text{sin 47}{}^\circ +\cos 77{}^\circ \\\\=\text{sin 47}{}^\circ +\cos \left( {90{}^\circ -13{}^\circ } \right)\\\\=\text{sin 47}{}^\circ +\sin 13{}^\circ \\\\=2\sin \displaystyle \frac{{\text{47}{}^\circ +13{}^\circ }}{2}\cos \displaystyle \frac{{\text{47}{}^\circ -13{}^\circ }}{2}\\\\=2\sin \text{30}{}^\circ \cos \text{17}{}^\circ \\\\=2\left( {\displaystyle \frac{1}{2}} \right)\cos \text{17}{}^\circ \\\\=\cos \text{17}{}^\circ \\\\\text{(b)}\\\ \ \ \cos 80{}^\circ +\cos 40{}^\circ -\cos 20{}^\circ \\\\=2\cos \displaystyle \frac{{\text{80}{}^\circ +40{}^\circ }}{2}\cos \displaystyle \frac{{\text{80}{}^\circ -40{}^\circ }}{2}-\cos 20{}^\circ \\\\=2\cos \text{60}{}^\circ \cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=2\left( {\displaystyle \frac{1}{2}} \right)\cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=\cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=0\end{array}$

6.        Prove that $ \displaystyle {{(\sin \alpha -\sin \beta )}^{2}}+{{(\cos \alpha -\cos \beta )}^{2}}=4{{\sin }^{2}}\left( {\frac{{\alpha -\beta }}{2}} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {\sin \alpha -\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha -\cos \beta } \right)}^{2}}\\\\=\ \ {{\cos }^{2}}\alpha -2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\ \ \ \ +\ {{\sin }^{2}}\alpha -2\sin \alpha \sin \beta +{{\sin }^{2}}\beta \\\\=\ \ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\\ \ \ \ -2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\\\=\ \ 2-2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\=\ \ 2\left[ {1-\cos \left( {\alpha -\beta } \right)} \right]\\\\=\ \ 2\left[ {1-\cos 2\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=\ \ 2\left[ {1-\left( {1-2{{{\sin }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right)} \right]\\\\=\ \ 2\left[ {2{{{\sin }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=4{{\sin }^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)\end{array}$

7.        Prove that $ \displaystyle {{(\sin \alpha +\sin \beta )}^{2}}+{{(\cos \alpha +\cos \beta )}^{2}}=4{{\cos }^{2}}\left( {\frac{{\alpha -\beta }}{2}} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {\sin \alpha +\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta } \right)}^{2}}\\\\=\ \ {{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\ \ \ \ +\ {{\sin }^{2}}\alpha +2\sin \alpha \sin \beta +{{\sin }^{2}}\beta \\\\=\ \ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\\ \ \ \ +2\cos \alpha \cos \beta +2\sin \alpha \sin \beta \\\\=\ \ 2+2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\=\ \ 2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\=\ \ 2\left[ {1+\cos 2\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=\ \ 2\left[ {1+\left( {2{{{\cos }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)-1} \right)} \right]\\\\=\ \ 2\left[ {2{{{\cos }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=4{{\cos }^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)\end{array}$

8.        Prove that $ \displaystyle \frac{{\sin \theta +\sin 2\theta +\sin 4\theta +\sin 5\theta }}{{\cos \theta +\cos 2\theta +\cos 4\theta +\cos 5\theta }}=\tan 3\theta $

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{\sin \theta +\sin 2\theta +\sin 4\theta +\sin 5\theta }}{{\cos \theta +\cos 2\theta +\cos 4\theta +\cos 5\theta }}\\\\=\ \displaystyle \frac{{\sin 5\theta +\sin \theta +\sin 4\theta +\sin 2\theta }}{{\cos 5\theta +\cos \theta +\cos 4\theta +\cos 2\theta }}\\\\=\ \displaystyle \frac{{2\sin \displaystyle \frac{{5\theta +\theta }}{2}\cos \displaystyle \frac{{5\theta -\theta }}{2}+2\sin \displaystyle \frac{{4\theta +2\theta }}{2}\cos \displaystyle \frac{{4\theta -2\theta }}{2}}}{{2\cos \displaystyle \frac{{5\theta +\theta }}{2}\cos \displaystyle \frac{{5\theta -\theta }}{2}+2\cos \displaystyle \frac{{4\theta +2\theta }}{2}\cos \displaystyle \frac{{4\theta -2\theta }}{2}}}\\\\=\ \displaystyle \frac{{2\left( {\sin 3\theta \cos 2\theta +\sin 3\theta \cos \theta } \right)}}{{2\left( {\cos 3\theta \cos 2\theta +\cos 3\theta \cos \theta } \right)}}\\\\=\ \displaystyle \frac{{\sin 3\theta \left( {\cos 2\theta +\cos \theta } \right)}}{{\cos 3\theta \left( {\cos 2\theta +\cos \theta } \right)}}\\\\=\tan 3\theta \end{array}$

9.        If $\displaystyle \tan \alpha =\frac{m}{{m+1}}$ and $ \displaystyle \tan \beta =\frac{1}{{2m+1}}$, find the acute angle $ \displaystyle \theta$ such that $ \displaystyle \cot \theta =\tan \left( {\alpha +\beta } \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \alpha =\displaystyle \frac{m}{{m+1}},\tan \beta =\displaystyle \frac{1}{{2m+1}}\\\\\ \ \ \ \cot \theta =\tan \left( {\alpha +\beta } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{m}{{m+1}}+\displaystyle \frac{1}{{2m+1}}}}{{1-\displaystyle \frac{m}{{m+1}}\times \displaystyle \frac{1}{{2m+1}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2{{m}^{2}}+m+m+1}}{{\left( {m+1} \right)\left( {2m+1} \right)}}}}{{\displaystyle \frac{{2{{m}^{2}}+3m+1-m}}{{\left( {m+1} \right)\left( {2m+1} \right)}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2{{m}^{2}}+2m+1}}{{2{{m}^{2}}+2m+1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =1\\\\\therefore \ \ \cot \theta =\cot 45{}^\circ \\\\\therefore \ \ \theta =45{}^\circ \end{array}$

10.       Prove that

(a) $ \displaystyle \frac{{1+\sin 2\alpha -\cos 2\alpha }}{{1+\sin 2\alpha +\cos 2\alpha }}=\tan \alpha $

(b) $ \displaystyle \frac{{1+\sin 2\alpha +\cos 2\alpha }}{{1+\sin 2\alpha -\cos 2\alpha }}=\cot \alpha $

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$ \displaystyle \begin{array}{l}\text{(a)}\\\ \ \ \displaystyle \frac{{1+\sin 2\alpha -\cos 2\alpha }}{{1+\sin 2\alpha +\cos 2\alpha }}\\\\=\displaystyle \frac{{1+2\sin \alpha \cos \alpha -1+2{{{\sin }}^{2}}\alpha }}{{1+2\sin \alpha \cos \alpha +2{{{\cos }}^{2}}\alpha -1}}\\\\=\displaystyle \frac{{2\sin \alpha \left( {\sin \alpha +\cos \alpha } \right)}}{{2\cos \alpha \left( {\sin \alpha +\cos \alpha } \right)}}\\\\=\tan \alpha \\\\\text{(b)}\\\ \ \ \displaystyle \frac{{1+\sin 2\alpha +\cos 2\alpha }}{{1+\sin 2\alpha -\cos 2\alpha }}\\\\=\displaystyle \frac{{1+2\sin \alpha \cos \alpha +2{{{\cos }}^{2}}\alpha -1}}{{1+2\sin \alpha \cos \alpha -1+2{{{\sin }}^{2}}\alpha }}\\\\=\displaystyle \frac{{2\cos \alpha \left( {\sin \alpha +\cos \alpha } \right)}}{{2\sin \alpha \left( {\sin \alpha +\cos \alpha } \right)}}\\\\=\cot \alpha \end{array}$

11.       If $ \displaystyle \sec \theta + \tan \theta = 3$, where $ \displaystyle \theta$ lies in the first quadrant, then find the value of $ \displaystyle \cos \theta$.

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$ \displaystyle \begin{array}{l}\ \ \ \sec \theta +\tan \theta =3\ ---(1)\\\\\ \ \ \displaystyle \frac{{\left( {\sec \theta +\tan \theta } \right)\left( {\sec \theta -\tan \theta } \right)}}{{\sec \theta -\tan \theta }}=3\\\\\ \ \ \displaystyle \frac{{{{{\sec }}^{2}}\theta -{{{\tan }}^{2}}\theta }}{{\sec \theta -\tan \theta }}=3\\\\\ \ \ \displaystyle \frac{1}{{\sec \theta -\tan \theta }}=3\\\\\therefore \ \ \sec \theta -\tan \theta =\displaystyle \frac{1}{3}\ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\sec \theta =\displaystyle \frac{{10}}{3}\\\\\therefore \ \sec \theta =\displaystyle \frac{5}{3}\\\\\therefore \ \cos \theta =\displaystyle \frac{3}{5}\end{array}$

12.       If $ \displaystyle \operatorname{cosec}\theta -\cot \theta =\frac{1}{5}$, where $ \displaystyle \theta$ lies in the first quadrant, then find the value of $ \displaystyle \sin \theta$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \operatorname{cosec}\theta -\cot \theta =\displaystyle \frac{1}{5}\ \ \ ---(1)\\\\\therefore \ \ \displaystyle \frac{{\left( {\operatorname{cosec}\theta -\cot \theta } \right)\left( {\operatorname{cosec}\theta +\cot \theta } \right)}}{{\left( {\operatorname{cosec}\theta +\cot \theta } \right)}}=\displaystyle \frac{1}{5}\\\\\therefore \ \ \displaystyle \frac{{{{{\operatorname{cosec}}}^{2}}\theta -{{{\cot }}^{2}}\theta }}{{\operatorname{cosec}\theta +\cot \theta }}=\displaystyle \frac{1}{5}\\\\\therefore \ \ \displaystyle \frac{1}{{\operatorname{cosec}\theta +\cot \theta }}=\displaystyle \frac{1}{5}\\\ \ \ \left[ {\because 1+{{{\cot }}^{2}}\theta ={{{\operatorname{cosec}}}^{2}}\theta } \right]\\\\\therefore \ \ \operatorname{cosec}\theta +\cot \theta =5\ \ \ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\operatorname{cosec}\theta =\displaystyle \frac{{26}}{5}\\\\\therefore \ \operatorname{cosec}\theta =\displaystyle \frac{{13}}{5}\\\\\therefore \ \sin \theta =\displaystyle \frac{5}{{13}}\end{array}$

13.       If $ \displaystyle \cos \theta +\sin \theta =\sqrt{2}\cos \theta $, then prove that $ \displaystyle \cos \theta -\sin \theta =\sqrt{2}\sin \theta $.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos \theta +\sin \theta =\sqrt{2}\cos \theta \\\\\therefore \ \ {{\left( {\cos \theta +\sin \theta } \right)}^{2}}=2{{\cos }^{2}}\theta \ \\\\\ \ \ \ {{\cos }^{2}}\theta +2\sin \theta \cos \theta +{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta \ \\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta -{{\sin }^{2}}\theta =0\\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta -{{\sin }^{2}}\theta +2{{\sin }^{2}}\theta =2{{\sin }^{2}}\theta \\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta +{{\sin }^{2}}\theta =2{{\sin }^{2}}\theta \\\\\therefore \ \ {{\left( {\cos \theta -\sin \theta } \right)}^{2}}=2{{\sin }^{2}}\theta \\\\\therefore \ \ \cos \theta -\sin \theta =\sqrt{2}\sin \theta \end{array}$

14.       If $ \displaystyle \sin \left( {\alpha -\beta } \right)=\cos \left( {\alpha +\beta } \right)=\frac{1}{2}$, find the positive acute angles $ \displaystyle \alpha$ and $ \displaystyle \beta$.

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$ \displaystyle \begin{array}{l}\ \ \ \sin \left( {\alpha -\beta } \right)=\displaystyle \frac{1}{2}\\\\\therefore \ \alpha -\beta =30{}^\circ \,\ ---(1)\\\\\ \ \ \cos \left( {\alpha +\beta } \right)=\displaystyle \frac{1}{2}\\\\\therefore \ \alpha +\beta =60{}^\circ \ \ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\alpha =90{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha =45{}^\circ \\\\\ \ \ (2)-(1)\Rightarrow 2\beta =30{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta =15{}^\circ \end{array}$

15.       Prove that $ \displaystyle \frac{{2\sin \theta }}{{1+\sin \theta +\cos \theta }}+\frac{{\cos \theta }}{{1+\sin \theta }}=1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{2\sin \theta }}{{1+\sin \theta +\cos \theta }}+\displaystyle \frac{{\cos \theta }}{{1+\sin \theta }}\\\\=\ \ \displaystyle \frac{{2\sin \theta \left( {1+\sin \theta } \right)+\cos \theta \left( {1+\sin \theta +\cos \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{2\sin \theta +2{{{\sin }}^{2}}\theta +\cos \theta +\sin \theta \cos \theta +{{{\cos }}^{2}}\theta }}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta +\sin \theta +\sin \theta +{{{\sin }}^{2}}\theta +\cos \theta +\sin \theta \cos \theta }}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{\left( {1+\sin \theta } \right)+\sin \theta \left( {1+\sin \theta } \right)+\cos \theta \left( {1+\sin \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{\left( {1+\sin \theta } \right)\left( {1+\sin \theta +\cos \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=1\end{array}$

16.      If $ \displaystyle \tan \theta =\frac{4}{5}$, where $ \displaystyle \theta$ lies in the first quadrant, find the exact value of $ \displaystyle \frac{{5\sin \theta -3\cos \theta }}{{\sin \theta +2\cos \theta }}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \theta =\displaystyle \frac{4}{5},\\\\\ \ \ \ \displaystyle \frac{{5\sin \theta -3\cos \theta }}{{\sin \theta +2\cos \theta }}\\\\=\ \displaystyle \frac{{5\displaystyle \frac{{\sin \theta }}{{\cos \theta }}-3\displaystyle \frac{{\cos \theta }}{{\cos \theta }}}}{{\displaystyle \frac{{\sin \theta }}{{\cos \theta }}+2\displaystyle \frac{{\cos \theta }}{{\cos \theta }}}}\\\\=\ \displaystyle \frac{{5\tan \theta -3}}{{\tan \theta +2}}\\\\=\ \displaystyle \frac{{5\left( {\displaystyle \frac{4}{5}} \right)-3}}{{\left( {\displaystyle \frac{4}{5}} \right)+2}}\\\\=\displaystyle \frac{1}{{\left( {\displaystyle \frac{{14}}{5}} \right)}}\\\\=\displaystyle \frac{5}{{14}}\end{array}$

17.       If $ \displaystyle \tan \theta +\sin \theta =m$ and $ \displaystyle \tan \theta -\sin \theta =n$, show that $ \displaystyle {{m}^{2}}-{{n}^{2}}=4\sqrt{{mn}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \tan \theta +\sin \theta =m,\ \\\\\ \ \ \tan \theta -\sin \theta =n\\\\\therefore \ \ m+n=2\tan \theta \\\\\therefore \ \ m-n=2\sin \theta \\\\\therefore \ \ \left( {m+n} \right)\left( {m-n} \right)=2\tan \theta \times 2\sin \theta \\\\\therefore \ \ {{m}^{2}}-{{n}^{2}}=4\sin \theta \tan \theta \\\\\ \ \ \ mn=\left( {\tan \theta +\sin \theta } \right)\left( {\tan \theta -\sin \theta } \right)\\\\\ \ \ \ mn={{\tan }^{2}}\theta -{{\sin }^{2}}\theta \\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}-{{\sin }^{2}}\theta \\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta -{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta \left( {1-{{{\cos }}^{2}}\theta } \right)}}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta {{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn={{\sin }^{2}}\theta {{\tan }^{2}}\theta \\\\\therefore \ \ \sqrt{{mn}}=\sin \theta \tan \theta \\\\\therefore \ \ 4\sqrt{{mn}}=4\sin \theta \tan \theta \\\\\therefore \ \ {{m}^{2}}-{{n}^{2}}=4\sqrt{{mn}}\end{array}$

18.       Without using tables, prove that $ \displaystyle \cos 105{}^\circ +\cos 15{}^\circ =\sin 75{}^\circ -\sin 15{}^\circ $.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos 105{}^\circ +\cos 15{}^\circ \\\\=\,\ 2\cos \displaystyle \frac{{105{}^\circ +15{}^\circ }}{2}\cos \displaystyle \frac{{105{}^\circ -15{}^\circ }}{2}\\\\=\,\ 2\cos 60{}^\circ \cos 45{}^\circ \\\\=\,\ 2\left( {\displaystyle \frac{1}{2}} \right)\displaystyle \frac{{\sqrt{2}}}{2}\\\\=\,\ \displaystyle \frac{{\sqrt{2}}}{2}\\\\\ \ \ \ \sin 75{}^\circ -\sin 15{}^\circ \\\\=\,\ 2\cos \displaystyle \frac{{75{}^\circ +15{}^\circ }}{2}\sin \displaystyle \frac{{75{}^\circ -15{}^\circ }}{2}\\\\=\,\ 2\cos 45{}^\circ \sin 30{}^\circ \\\\=\,\ 2\left( {\displaystyle \frac{{\sqrt{2}}}{2}} \right)\displaystyle \frac{1}{2}\\\\=\,\ \displaystyle \frac{{\sqrt{2}}}{2}\\\\\therefore \ \cos 105{}^\circ +\cos 15{}^\circ =\sin 75{}^\circ -\sin 15{}^\circ \end{array}$

19.       Prove that $ \displaystyle \sin \theta +\cos \theta =\sqrt{2}\sin \left( {\theta +45{}^\circ } \right)=\sqrt{2}\cos \left( {\theta -45{}^\circ } \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \theta +\cos \theta \\\\=\,\sqrt{2}\ \left( {\displaystyle \frac{1}{{\sqrt{2}}}\sin \theta +\displaystyle \frac{1}{{\sqrt{2}}}\cos \theta } \right)\\\\=\,\sqrt{2}\ \left( {\sin \theta \cos 45{}^\circ +\cos \theta \sin 45{}^\circ } \right)\\\ \ \ \ \left[ {\because \sin 45{}^\circ =\cos 45{}^\circ =\displaystyle \frac{1}{{\sqrt{2}}}} \right]\\\\=\,\sqrt{2}\ \sin \left( {\theta +45{}^\circ } \right)\\\\\ \ \ \ \sin \theta +\cos \theta \\\\=\,\sqrt{2}\ \left( {\displaystyle \frac{1}{{\sqrt{2}}}\sin \theta +\displaystyle \frac{1}{{\sqrt{2}}}\cos \theta } \right)\\\\=\,\sqrt{2}\ \left( {\sin \theta \sin 45{}^\circ +\cos \theta \cos 45{}^\circ } \right)\\\ \ \ \ \left[ {\because \sin 45{}^\circ =\cos 45{}^\circ =\displaystyle \frac{1}{{\sqrt{2}}}} \right]\\\\=\,\sqrt{2}\ \left( {\cos \theta \cos 45{}^\circ +\sin \theta \sin 45{}^\circ } \right)\\\\=\,\sqrt{2}\ \cos \left( {\theta -45{}^\circ } \right)\end{array}$

20.       In $ \displaystyle \vartriangle ABC$, if $ \displaystyle \cot A+\cot B+\cot C=\sqrt{3}$, prove that $ \displaystyle \vartriangle ABC$ is an equilateral triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\\\\\ \ \ \ A+B+C=180{}^\circ \\\\\ \ \ \ A+B=180{}^\circ -C\\\\\ \ \ \ \tan \left( {A+B} \right)=\tan \left( {180{}^\circ -C} \right)\\\\\ \ \ \ \displaystyle \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=-\tan C\\\\\ \ \ \ \displaystyle \frac{{\displaystyle \frac{1}{{\cot A}}+\displaystyle \frac{1}{{\cot B}}}}{{1-\displaystyle \frac{1}{{\cot A\cot B}}}}=-\displaystyle \frac{1}{{\cot C}}\\\\\ \ \ \ \displaystyle \frac{{\displaystyle \frac{{\cot A+\cot B}}{{\cot A\cot B}}}}{{\displaystyle \frac{{\cot A\cot B-1}}{{\cot A\cot B}}}}=-\displaystyle \frac{1}{{\cot C}}\\\\\ \ \ \ \displaystyle \frac{{\cot A+\cot B}}{{\cot A\cot B-1}}=-\displaystyle \frac{1}{{\cot C}}\\\\\therefore \ \ \cot A\cot C+\cot B\cot C=1-\cot A\cot B\\\\\therefore \ \ \cot A\cot B+\cot B\cot C+\cot A\cot C=1\ \\\\\ \ \ \ \text{Let}\ \cot A=x,\ \cot B=y,\ \cot C=z\\\\\therefore \ \ xy+yz+xz=1\ ---(1)\\\\\ \ \ \ x+y+z=\sqrt{3}\,\ \left[ {\because given} \right]\\\\\therefore \ \ {{\left( {x+y+z} \right)}^{2}}=3\\\\\ \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( {xy+yz+xz} \right)=3\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(1)=3\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1=0\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz=0\ ---(2)\\\\\therefore \ \ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-2xy-2yz-2xz=0\\\\\therefore \ \left( {{{x}^{2}}-2xy+{{y}^{2}}} \right)+\left( {{{y}^{2}}-2yz+{{z}^{2}}} \right)+\left( {{{x}^{2}}-2xz+{{z}^{2}}} \right)=0\\\\\therefore \ {{\left( {x-y} \right)}^{2}}+{{\left( {y-z} \right)}^{2}}+{{\left( {x-z} \right)}^{2}}=0\\\\\therefore \ \ x-y=0\Rightarrow x=y\\\\\ \ \ y-z=0\Rightarrow y=z\\\\\ \ \ x-z=0\Rightarrow x=z\\\\\therefore \ \ x=y=z\Rightarrow \cot A=\cot B=\cot C\\\\\therefore \ \ A=B=C\\\\\therefore \ \ \vartriangle ABC\ \text{is an equilateral triangle}\text{.}\ \end{array}$

Target Mathematics : Volume 1 & 2

 

စာအုပ်အမည် - Terget Mathematics Vol (1) Algebra & Probability
အတန်း - တက္ကသိုလ်ဝင်တန်း (grade 11)
စီစဉ်ရေးသားသူ - ဆရာ သူရိန်မင်း
တန်းဘိုး - 2500 ကျပ်

Target Mathematics Vol-1 သည် Grade (11) Mathematics ပြဌာန်းသင်ရိုးပါ
Chapter (1) Functions
Chapter (2) The Remainder Theorem and The Factor Theorem
Chapter (3) The Binomial Theorem
Chapter (4) Inequations
Chapter (5) Sequences and Series
Chapter (6) Matrices
Chapter (7) Introduction to Probability
အခန်း (၇)ခန်း နှင့်သက်ဆိုင်သော 

- ပြဌာန်းစာအုပ်မှ ပုစ္ဆာကောင်းများ

- IGCSE, IB, SAT, IIT JEE စသော နိုင်ငံတကာ တက္ကသိုလ်ဝင် စာမေးပွဲများတွင် စစ်ဆေးခဲ့သော ပုစ္ဆာကောင်းများကို အဖြေနှင့်တကွ စုဆည်းဖော်ပြထားပါသည်။

ကျောင်းသားများ အလွယ်တကူလေ့လာနိုင်စေရန် လိုအပ်သော နေရာများတွင် illustrative diagram များကိုလည်း ရေးဆွဲဖော်ပြထားပါသည်။

 

- စာအုပ်အမည် - Terget Mathematics Vol (2) Geometry, Vector, Trigonometry & Calculus
- အတန်း - တက္ကသိုလ်ဝင်တန်း (grade 11)
- စီစဉ်ရေးသားသူ - ဆရာ သူရိန်မင်း
- တန်းဘိုး - 2500 ကျပ်

Target Mathematics Vol-2 သည် Grade (11) Mathematics ပြဌာန်းသင်ရိုးပါ
Chapter (8) Circles
Chapter (9) Area of Similar Triangles
Chapter (10) Introduction to Vectors and Transformation Geometry
Chapter (11) Trigonometry
Chapter (12) Calculus
အခန်း (5)ခန်း နှင့်သက်ဆိုင်သော ပုစ္ဆာကောင်းများကို အဖြေနှင့်တကွ စုဆည်းဖော်ပြထားပါသည်။

ကျောင်းသားများ အလွယ်တကူလေ့လာနိုင်စေရန် လိုအပ်သော နေရာများတွင် illustrative diagram များကိုလည်း ရေးဆွဲဖော်ပြထားပါသည်။ သင်ရိုးပြင်ပ ပုစ္ဆာအလှအပများကိုလည်း ထည့်သွင်းပေးထားရာ ကျောင်းသာများအတွက် များစွာ အထောက်အကူ ဖြစ်စေလိမ့်မည်ဟု ယုံကြည်မိပါသည်။

မှာယူလိုပါက....

ဆရာ လှ (ရန်ကုန်) - 09-428121309

ဆရာ တင်နိုင်ဝင်း (မန္တလေး) 09-43133420

ဒေါ်ချမ်းမြေ့ခင် (ရန်ကုန်) 09 - 790177054 ထို့ ထံ ဆက်သွယ် မှာယူနိုင်ပါသည်။

အောက်ပါ ပုံစံ တွင်လဲ ပုံစံဖြည့်၍ မှာယူနိုင်ပါသည် ။

Sequences and Series

Sequence

A sequence is a function whose domain is either the set of all or part of the natural numbers. The values of the function (images) are called the terms of the of the sequence. 
Sequence ဆိုတာ Function တစ်ခုဖြစ်ပါတယ် ၎င်း၏ domain မှာ set of natural numbers (သဘာဝကိန်းများ ပါဝင်သော အစု) ဖြစ်ပါသည်။ Image များကိုတော့ term လို့ ခေါ်ပါတယ်။

ပုံတွင်

1 ၏ image သည် 1 ဖြစ်လို့ ⇒ f(1) = 1
2 ၏ image သည် 4 ဖြစ်လို့ ⇒ f(2) = 4
3 ၏ image သည် 9 ဖြစ်လို့ ⇒ f(3) = 9 
4 ၏ image သည် 16 ဖြစ်လို့ ⇒ f(4) = 16
⋮  

n ၏ image သည် n² ဖြစ်လို့ ⇒ f(n) = n² သတ်မှတ်နိုင်ပါတယ်။ အထက်ပါ သတ်မှတ်ချက်ဟာ function ကို လေ့လာခဲ့စဉ်က သတ်မှတ်သော functional notation ဖြစ်ပါတယ်။ Sequence မှာတော့ functional notation အစား အခြား notation ကို ပြောင်းသုံးပါမယ်။ ဒါ့ကြောင့်

$ \displaystyle f(1)$ အစား $ \displaystyle u_1 = 1^{\text{st}}\ \text{term}$

$ \displaystyle f(2)$ အစား $ \displaystyle u_2= 2^{\text{nd}}\ \text{term}$

$ \displaystyle f(3)$ အစား $ \displaystyle u_3= 3^{\text{rd}}\ \text{term}$

 ⋮
 $ \displaystyle f(n)$ အစား $ \displaystyle u_n= n^{\text{th}}\ \text{term}$ လို့ ပြောင်းသုံးပါမယ်။

The value of the function corresponding to the number n of the domain is called the n^th term or the general term of the sequence.

Domain ထဲရှိ အစုဝင် n နှင့် ဆက်စက်နေသော image ($ \displaystyle u_n$) ကို n^th term ဟု ခေါ်ပါတယ်။ ၎င်းကို sequence တစ်ခု ၏ general term လို့ လဲ သတ်မှတ်ပါတယ်။ Sequence ဖြစ်ပေါ်လာပုံ သေနည်း (Rule of Formation) လို့ ပြောနိုင်ပါတယ်။

Progression

When terms of a sequence are written under specific conditions, then the sequence is called a progression.

sequence တစ်ခုကို ဖြစ်ပေါ်လာပုံနည်း စနစ်တစ်ခု နဲ့ တည်ဆောက်ထားရင် ... တစ်နည်း general term တည်ဆောက်နိုင်နိုင်ရင် progression လို့ ခေါ်ပါတယ်။

progression တိုင်းက sequence များ ဖြစ်ကြပေမယ့် sequence တိုင်းကတော့ progression တွေ မဟုတ်ကြပါဘူး...

(1) သဘာဝကိန်းများ
(2) နှစ်ထပ်ကိန်းတိများ
(3) သုဒ္ဓကိန်းများ 

အထက်ပါသတ်မှတ်ချက်တွေဟာ sequence တွေ ဖြစ်တယ်လို့ ဆိုနိုင်ပါတယ်.. ဒါပေမယ့် သဘာဝကိန်းနဲ့ နှစ်ထပ်ကိန်းတိများ ဆိုတဲ့ သတ်မှတ်ချက်အတွက် general term (formula) ရှာလို့ရပေမယ့် သုဒ္ဓကိန်း တွေအတွက်တော့ general term ရှာလို့မရပါဘူး ဒါ့ကြောင့် သုဒ္ဓကိန်းများ ဖြစ်တဲ့..

2, 3, 5, 7, 11, ... ဆိုတာ sequnce တစ်ခုလို့ ဆိုနိုင်သော်လည်း progression တစ်ခုတော့ မဟုတ်ပါဘူး။ အလားတူးပါပဲ.. 1976 မှ 2019 အထိ ချဲဂဏန်း (ပေါက်ဂဏန်းများ) ဆိုတာ sequence တစ်ခု ဖြစ်ပါတယ်။ ဒါပေမယ့် progression တော့ မဟုတ်ပါဘူး။ ဘယ်နှစ်ရဲ့ ဘယ်အကြိမ်မှာ ဘာဂဏန်း ထွက်လာမယ်လို့ အတိအကျ မသတ်မှတ်နိုင်လို့ပါပဲ။ 

Binomial Theorem : Practice Problems

1.        The first three terms in the expansion of $ \displaystyle (1 + ax)^n$ are $ \displaystyle 1 + 12x + 64x^2$. Find $ \displaystyle n$ and $ \displaystyle a$.

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$ \displaystyle \begin{array}{l}\ \ \ {{(1+ax)}^{n}}=1+12x+64{{x}^{2}}+...\\\\\ \ \ {}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}+{}^{n}{{C}_{1}}{{\left( 1 \right)}^{{n-1}}}\left( {ax} \right)+{}^{n}{{C}_{2}}{{\left( 1 \right)}^{{n-2}}}{{\left( {ax} \right)}^{2}}+...=1+12x+64{{x}^{2}}+...\\\\\ \ \ 1+nax+\displaystyle \frac{{n\left( {n-1} \right)}}{2}{{a}^{2}}{{x}^{2}}+...=1+12x+64{{x}^{2}}+...\\\\\therefore \ na=12\Rightarrow a=\displaystyle \frac{{12}}{n}\\\\\ \ \displaystyle \frac{{n\left( {n-1} \right)}}{2}{{a}^{2}}=64\\\\\therefore \displaystyle \frac{{n\left( {n-1} \right)}}{2}{{\left( {\displaystyle \frac{{12}}{n}} \right)}^{2}}=64\\\\\therefore \displaystyle \frac{{72\left( {n-1} \right)}}{n}=64\\\\\therefore \ 72n-72=64n\Rightarrow n=9\\\\\therefore a=\displaystyle \frac{{12}}{9}=\displaystyle \frac{4}{3}\end{array}$

2.        Find the coefficient of $ \displaystyle x^3$ in the expansion of $ \displaystyle (1 + x + x^2)^3$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(1+x+{{x}^{2}})}^{3}}\\\\={{\left( {1+\left( {x+{{x}^{2}}} \right)} \right)}^{3}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\=1+3\left( {x+{{x}^{2}}} \right)+3{{\left( {x+{{x}^{2}}} \right)}^{2}}+{{\left( {x+{{x}^{2}}} \right)}^{3}}\\\\=1+3\left( {x+{{x}^{2}}} \right)+3\left( {{{x}^{2}}+2{{x}^{3}}+{{x}^{4}}} \right)+{{\left( {{{x}^{3}}+3{{x}^{4}}+3{{x}^{5}}+{{x}^{6}}} \right)}^{3}}\\\\\therefore \ \text{Coefficient of}\ {{x}^{3}}=3\left( 2 \right)+1=7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

3.        Find the value of $ \displaystyle {{\left( {{{a}^{2}}+\sqrt{{{{a}^{2}}-1}}} \right)}^{4}}+{{\left( {{{a}^{2}}-\sqrt{{{{a}^{2}}-1}}} \right)}^{4}}$.

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$ \displaystyle \begin{array}{l}\ \ \ {{\left( {x+y} \right)}^{4}}={{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}\\\\\ \ \ {{\left( {x-y} \right)}^{4}}={{x}^{4}}-4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}-4x{{y}^{3}}+{{y}^{4}}\\\\\therefore {{\left( {x+y} \right)}^{4}}+{{\left( {x-y} \right)}^{4}}\\\\=2{{x}^{4}}+12{{x}^{2}}{{y}^{2}}+2{{y}^{4}}\\\\\ \ \ \text{Taking}\ x={{a}^{2}}\ \text{and}\ y=\sqrt{{{{a}^{2}}-1}}\\\\\ \ \ {{\left( {{{a}^{2}}+\sqrt{{{{a}^{2}}-1}}} \right)}^{4}}+{{\left( {{{a}^{2}}-\sqrt{{{{a}^{2}}-1}}} \right)}^{4}}\\\\=2{{\left( {{{a}^{2}}} \right)}^{4}}+12{{\left( {{{a}^{2}}} \right)}^{2}}{{\left( {\sqrt{{{{a}^{2}}-1}}} \right)}^{2}}+2{{\left( {\sqrt{{{{a}^{2}}-1}}} \right)}^{4}}\\\\=2{{a}^{8}}+12{{a}^{4}}\left( {{{a}^{2}}-1} \right)+2\left( {{{a}^{4}}-2{{a}^{2}}+1} \right)\\\\=2{{a}^{8}}+12{{a}^{6}}-12{{a}^{4}}+2{{a}^{4}}-4{{a}^{2}}+2\\\\=2{{a}^{8}}+12{{a}^{6}}-10{{a}^{4}}-4{{a}^{2}}+2\\\\=2\left( {{{a}^{8}}+6{{a}^{6}}-5{{a}^{4}}-2{{a}^{2}}+1} \right)\end{array}$

4.        Find the coefficient of $ \displaystyle x^6y^3$ in the expansion of $ \displaystyle (x + 2y)^9$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {x+2y} \right)}^{9}}\\\\=\ \ {}^{9}{{C}_{r}}{{x}^{{9-r}}}{{\left( {2y} \right)}^{r}}\\\\=\ \ {}^{9}{{C}_{r}}{{2}^{r}}{{x}^{{9-r}}}{{y}^{r}}\\\\\ \ \ \ \text{For }{{x}^{6}}{{y}^{3}},\ r=3\\\\\therefore \ \ \text{Coefficient of }{{x}^{6}}{{y}^{3}}\\\\=\ \ {}^{9}{{C}_{3}}{{2}^{3}}\\\\=\ \ \displaystyle \frac{{9\times 8\times 7}}{{1\times 2\times 3}}\times {{2}^{3}}\\\\=672\end{array}$

5.        Find the term independent of $ \displaystyle x$ in the expansion of $ \displaystyle {{\left( {\frac{{3{{x}^{2}}}}{2}-\frac{1}{{3x}}} \right)}^{9}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {\displaystyle \frac{{3{{x}^{2}}}}{2}-\displaystyle \frac{1}{{3x}}} \right)}^{9}}\\\\=\ \ {}^{9}{{C}_{r}}{{\left( {\displaystyle \frac{{3{{x}^{2}}}}{2}} \right)}^{{9-r}}}{{\left( {-\displaystyle \frac{1}{{3x}}} \right)}^{r}}\\\\=\ \ {}^{9}{{C}_{r}}{{\left( {\displaystyle \frac{3}{2}} \right)}^{{9-r}}}{{\left( {-\displaystyle \frac{1}{3}} \right)}^{r}}{{x}^{{18-3r}}}\\\\\ \ \ \ \text{For the term independent of }x,\ \\\\\ \ \ \ 18-3r=0\Rightarrow r=6\\\\\therefore \ \ \text{The term independent of }x\\\\=\ \ {}^{9}{{C}_{6}}{{\left( {\displaystyle \frac{3}{2}} \right)}^{3}}{{\left( {-\displaystyle \frac{1}{3}} \right)}^{6}}\\\\={}^{9}{{C}_{3}}{{\left( {\displaystyle \frac{3}{2}} \right)}^{3}}{{\left( {-\displaystyle \frac{1}{3}} \right)}^{6}}\,\ \ \left[ {{}^{9}{{C}_{6}}={}^{9}{{C}_{{9-6}}}={}^{9}{{C}_{3}}} \right]\\\\=\displaystyle \frac{{9\times 8\times 7}}{{1\times 2\times 3}}\times {{\left( {\displaystyle \frac{3}{2}} \right)}^{3}}\times {{\left( {-\displaystyle \frac{1}{3}} \right)}^{6}}\\\\=\displaystyle \frac{7}{{18}}\end{array}$

6.        If the coefficients of $ \displaystyle x^7$ and $ \displaystyle x^{-7}$ in the expansion of $ \displaystyle {{\left( {ax+\frac{1}{{bx}}} \right)}^{{11}}}$ are equal, prove that $ \displaystyle ab = 1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}\ {{\left( {ax+\displaystyle \frac{1}{{bx}}} \right)}^{{11}}}\\\\=\ \ {}^{{11}}{{C}_{r}}{{\left( {ax} \right)}^{{11-r}}}{{\left( {\displaystyle \frac{1}{{bx}}} \right)}^{r}}\\\\=\ \ {}^{{11}}{{C}_{r}}{{\left( a \right)}^{{11-r}}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{r}}{{x}^{{11-2r}}}\\\\\ \ \ \ \text{For }{{x}^{7}},11-2r=7\Rightarrow r=2\ \\\\\therefore \ \ \text{Coefficient of }{{x}^{7}}={}^{{11}}{{C}_{2}}{{\left( a \right)}^{9}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{2}}\\\\\ \ \ \ \text{For }{{x}^{{-7}}},11-2r=-7\Rightarrow r=9\ \\\\\therefore \ \ \text{Coefficient of }{{x}^{{-7}}}={}^{{11}}{{C}_{9}}{{\left( a \right)}^{2}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{9}}\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ {}^{{11}}{{C}_{2}}{{\left( a \right)}^{9}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{2}}=\ {}^{{11}}{{C}_{9}}{{\left( a \right)}^{2}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{9}}\\\\\ \ \ \ {}^{{11}}{{C}_{2}}{{\left( a \right)}^{9}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{2}}=\ {}^{{11}}{{C}_{2}}{{\left( a \right)}^{2}}{{\left( {\displaystyle \frac{1}{b}} \right)}^{9}}\\\ \ \ \ \left[ {\because {}^{{11}}{{C}_{9}}={}^{{11}}{{C}_{{11-9}}}={}^{{11}}{{C}_{2}}} \right]\\\\\therefore {{a}^{7}}=\displaystyle \frac{1}{{{{b}^{7}}}}\\\\\therefore {{a}^{7}}{{b}^{7}}=1\Rightarrow ab=1\end{array}$

7.        In the expansion of $ \displaystyle (1 + x)^{43}$, the coefficients of $ \displaystyle (2p + 1)^{\text{th}}$ and $ \displaystyle (p + 2)^{\text{th}}$ terms are equal where $ \displaystyle p>1$, find the value of $ \displaystyle p$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion of}\ {{\left( {1+x} \right)}^{{43}}}\\\\=\ \ {}^{{43}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{\left( {2p+1} \right)}^{{\text{th}}}}\text{ term}\ ={}^{{43}}{{C}_{{2p}}}{{x}^{{2p}}}\\\\\ \ \ \ {{\left( {p+2} \right)}^{{\text{th}}}}\text{ term}\ ={{\left[ {(p+1)+1} \right]}^{{\text{th}}}}\text{ term}\ ={}^{{43}}{{C}_{{p+1}}}{{x}^{{p+1}}}\ \\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ {}^{{43}}{{C}_{{2p}}}=\ {}^{{43}}{{C}_{{p+1}}}\Rightarrow 2p=p+1\Rightarrow p=1\\\\\ \ \ \ \text{Since}\ p>1,p=1\ \text{is impossible}\text{.}\\\\\ \ \ \ \text{But}\ {}^{n}{{C}_{r}}={}^{n}{{C}_{{n-r}}},\\\\\ \ \ \ {}^{{43}}{{C}_{{2p}}}=\ {}^{{43}}{{C}_{{43-(p+1)}}}\\\\\therefore \ \ \ 2p=42-p\Rightarrow p=14\end{array}$

8.        If the $ \displaystyle 21^{\text{st}}$ and $ \displaystyle 22^{\text{nd}}$ terms in the expansion of $ \displaystyle (1 + a)^{44}$ are equal then find the value of $ \displaystyle a$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}\ {{(1+a)}^{{44}}}={}^{{44}}{{C}_{r}}{{a}^{r}}\\\\\therefore \ \ {{21}^{{\text{st}}}}\text{ term}\ ={{(20+1)}^{{\text{th}}}}\text{ term}\ ={}^{{44}}{{C}_{{20}}}{{a}^{{20}}}\\\\\ \ \ \ {{22}^{{\text{nd}}}}\text{ term}\ ={{(21+1)}^{{\text{th}}}}\text{ term}\ ={}^{{44}}{{C}_{{21}}}{{a}^{{21}}}\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ \ {{21}^{{\text{st}}}}\text{ term}\ =\ {{22}^{{\text{nd}}}}\text{ term}\\\\\therefore \ \ \ {}^{{44}}{{C}_{{20}}}{{a}^{{20}}}={}^{{44}}{{C}_{{21}}}{{a}^{{21}}}\\\\\therefore \ \ \ {}^{{44}}{{C}_{{20}}}={}^{{44}}{{C}_{{20}}}\times \displaystyle \frac{{24}}{{21}}\times a\\\\\therefore \ \ \ a=\displaystyle \frac{{21}}{{24}}\end{array}$

9.        Find $ \displaystyle n$, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $ \displaystyle {{\left( {\sqrt[4]{2}+\frac{1}{{\sqrt[4]{3}}}} \right)}^{n}}$ is $ \displaystyle \sqrt{6}:1$.

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$ \displaystyle \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}\ {{\left( {\sqrt[4]{2}+\frac{1}{{\sqrt[4]{3}}}} \right)}^{n}}$ $ \displaystyle \ \ \ \ ={}^{n}{{C}_{r}}{{\left( {\sqrt[4]{2}} \right)}^{{n-r}}}{{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)}^{r}}$ $ \displaystyle \therefore \ \ {{5}^{{\text{th}}}}\text{ term from the beginning}={}^{n}{{C}_{4}}{{\left( {\sqrt[4]{2}} \right)}^{{n-4}}}{{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)}^{4}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={}^{n}{{C}_{4}}{{\left( {\sqrt[4]{2}} \right)}^{{n-4}}}\left( {\frac{1}{3}} \right)$ $ \displaystyle \ \ \ \ {{5}^{{\text{th}}}}\text{ term from the end}={}^{n}{{C}_{{n-4}}}{{\left( {\sqrt[4]{2}} \right)}^{4}}{{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)}^{{n-4}}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{C}_{{n-4}}}\left( 2 \right){{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)}^{{n-4}}}$ $ \displaystyle \ \ \ \ \text{By the problem,}\ $ $ \displaystyle \frac{{{}^{n}{{C}_{4}}{{{\left( {\sqrt[4]{2}} \right)}}^{{n-4}}}\left( {\displaystyle \frac{1}{3}} \right)}}{{{}^{n}{{C}_{{n-4}}}\left( 2 \right){{{\left( {\displaystyle \frac{1}{{\sqrt[4]{3}}}} \right)}}^{{n-4}}}}}\ =\displaystyle \frac{{\sqrt{6}}}{1}$ $ \displaystyle \ \ \ \ \ \text{Since}\ {}^{n}{{C}_{4}}={}^{n}{{C}_{{n-4}}},$ $ \displaystyle \ \ \ \ \frac{{{{{\left( 2 \right)}}^{{\frac{{n-4}}{4}}}}\left( {\frac{1}{3}} \right)}}{{\left( 2 \right){{{\left( {\frac{1}{3}} \right)}}^{{\frac{{n-4}}{4}}}}}}=\sqrt{6}$ $ \displaystyle \therefore \ \ \ {{\left( 2 \right)}^{{\frac{{n-4}}{4}-1}}}\times {{\left( { \displaystyle \frac{1}{3}} \right)}^{{1-\frac{{n-4}}{4}}}}=\sqrt{6}$ $ \displaystyle \therefore \ \ \ {{\left( 2 \right)}^{{\frac{{n-8}}{4}}}}\times {{\left( 3 \right)}^{{\frac{{n-8}}{4}}}}=\sqrt{6}$ $ \displaystyle \begin{array}{l}\therefore \ \ \ {{6}^{{\frac{{n-8}}{4}}}}={{6}^{{\frac{1}{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{n-8}}{4}=\frac{1}{2}\Rightarrow n=10\end{array}$

10.       If the coefficients of $ \displaystyle a^{k-1}$, $ \displaystyle a^k$, $ \displaystyle a^{k+1}$ in the binomial expansion of $ \displaystyle (1 + a)^n$ are in $ \displaystyle A.P.$, then prove that $ \displaystyle n^2 - n(4k + 1) + 4k^2 - 2 = 0$.

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$ \displaystyle \begin{array}{l}\ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {1+a} \right)}^{n}}=\ {}^{n}{{C}_{r}}{{a}^{n}}\\\\\therefore \ \text{Coefficient of}\ {{a}^{{k-1}}}={}^{n}{{C}_{{k-1}}}\\\\\ \ \ \text{Coefficient of}\ {{a}^{k}}={}^{n}{{C}_{k}}\\\\\ \ \ \text{Coefficient of}\ {{a}^{{k+1}}}={}^{n}{{C}_{{k+1}}}\\\\\ \ \ \text{By the problem,}\ \\\\\ \ \ {}^{n}{{C}_{{k-1}}},{}^{n}{{C}_{k}},{}^{n}{{C}_{{k+1}}}\ \text{are in A}\text{.P}\text{.}\\\\\therefore \ \ {}^{n}{{C}_{k}}-{}^{n}{{C}_{{k-1}}}={}^{n}{{C}_{{k+1}}}-{}^{n}{{C}_{k}}\\\\\therefore \ \ {}^{n}{{C}_{{k-1}}}+{}^{n}{{C}_{{k+1}}}=2{}^{n}{{C}_{k}}\\\\\therefore \ \ {}^{n}{{C}_{{k-1}}}+{}^{n}{{C}_{{k-1}}}\displaystyle \frac{{\left( {n-k} \right)\left( {n-k+1} \right)}}{{k\left( {k+1} \right)}}=2\cdot {}^{n}{{C}_{{k-1}}}\cdot \displaystyle \frac{{n-k+1}}{k}\\\\\therefore \ \ 1+\displaystyle \frac{{{{n}^{2}}-2kn+{{k}^{2}}+n-k}}{{k\left( {k+1} \right)}}=\displaystyle \frac{{2n-2k+2}}{k}\\\\\therefore \ \ {{k}^{2}}+k+{{n}^{2}}-2kn+{{k}^{2}}+n-k=2-2{{k}^{2}}+2n+2kn\\\\\therefore \ \ {{n}^{2}}-4kn-n+4{{k}^{2}}-2=0\\\\\therefore \ \ {{n}^{2}}-n\left( {4k+1} \right)+4{{k}^{2}}-2=0\end{array}$

11.      If the coefficients of three successive terms in the expansion of $ \displaystyle (1 + x)^n$ are in the ratio $ \displaystyle 1 : 3 : 5$, then find the value of $ \displaystyle n$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{n}}={{\ }^{n}}{{C}_{r}}{{x}^{n}}\\\\\therefore \ \ \text{Coefficient of}\ {{x}^{{r-1}}}={}^{n}{{C}_{{r-1}}}\\\\\ \ \ \ \text{Coefficient of}\ {{x}^{r}}={}^{n}{{C}_{r}}\\\\\ \ \ \ \text{Coefficient of}\ {{x}^{{r+1}}}={}^{n}{{C}_{{r+1}}}\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ {}^{n}{{C}_{{r-1}}}:{}^{n}{{C}_{r}}:{}^{n}{{C}_{{r+1}}}=1:3:5\\\\\therefore \ \ \displaystyle \frac{{{}^{n}{{C}_{{r-1}}}}}{{{}^{n}{{C}_{r}}}}=\displaystyle \frac{1}{3}\\\\\ \ \ \displaystyle \frac{{{}^{n}{{C}_{{r-1}}}}}{{{}^{n}{{C}_{{r-1}}}\displaystyle \frac{{n-r+1}}{r}}}=\displaystyle \frac{1}{3}\\\\\therefore \ \ \displaystyle \frac{r}{{n-r+1}}=\displaystyle \frac{1}{3}\\\\\therefore \ \ n=4r-1\\\\\ \ \ \text{Again,}\ \ \displaystyle \frac{{{}^{n}{{C}_{r}}}}{{{}^{n}{{C}_{{r+1}}}}}=\displaystyle \frac{3}{5}\\\\\ \ \ \displaystyle \frac{{{{C}_{r}}}}{{{}^{n}{{C}_{r}}\displaystyle \frac{{n-r}}{{r+1}}}}=\displaystyle \frac{1}{3}\\\\\therefore \ \ \displaystyle \frac{{r+1}}{{n-r}}=\displaystyle \frac{3}{5}\\\\\therefore \ \ \displaystyle \frac{{r+1}}{{4r-1-r}}=\displaystyle \frac{3}{5}\\\\\therefore \ \ r=2\Rightarrow n=7\end{array}$

12.       The second, third and fourth terms in the expansion of $ \displaystyle (x + a)^n$ are $ \displaystyle 240, 720$ and $ \displaystyle 1080$ respectively. Find the values of $ \displaystyle x, a$ and $ \displaystyle n$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {x+a} \right)}^{n}}={{\ }^{n}}{{C}_{r}}{{a}^{r}}{{x}^{{n-r}}}\\\\\therefore \ \ {{2}^{{\text{nd}}}}\text{ term}={{\left( {1+1} \right)}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{1}}a{{x}^{{n-1}}}\\\\\ \ \ \ {{3}^{{\text{rd}}}}\text{ term}={{\left( {2+1} \right)}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{2}}{{a}^{2}}{{x}^{{n-2}}}\\\\\ \ \ \ {{4}^{{\text{th}}}}\text{ term}={{\left( {3+1} \right)}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{3}}{{a}^{3}}{{x}^{{n-3}}}\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ {}^{n}{{C}_{1}}a{{x}^{{n-1}}}=240\ \ \ \ ---(1)\\\\\ \ \ \ {}^{n}{{C}_{2}}{{a}^{2}}{{x}^{{n-2}}}=720\ \ \ \ ---(2)\\\\\ \ \ \ {}^{n}{{C}_{3}}{{a}^{3}}{{x}^{{n-3}}}=1080\ \ \ \ ---(3)\\\\\ \ \ \ (2)\div (1)\Rightarrow \displaystyle \frac{{{}^{n}{{C}_{2}}{{a}^{2}}{{x}^{{n-2}}}}}{{{}^{n}{{C}_{1}}a{{x}^{{n-1}}}}}=\displaystyle \frac{{720}}{{240}}\\\\\therefore \ \ \displaystyle \frac{{{}^{n}{{C}_{1}}\displaystyle \frac{{n-1}}{2}}}{{{}^{n}{{C}_{1}}}}\displaystyle \frac{a}{x}=3\\\\\therefore \ \ \displaystyle \frac{a}{x}=\displaystyle \frac{6}{{n-1}}\\\\\ \ \ \ (3)\div (2)\Rightarrow \displaystyle \frac{{{}^{n}{{C}_{3}}{{a}^{3}}{{x}^{{n-3}}}}}{{{}^{n}{{C}_{2}}{{a}^{2}}{{x}^{{n-2}}}}}=\displaystyle \frac{{1080}}{{720}}\\\\\therefore \ \ \ \displaystyle \frac{{{}^{n}{{C}_{2}}\displaystyle \frac{{n-2}}{3}}}{{{}^{n}{{C}_{2}}}}\displaystyle \frac{a}{x}=\displaystyle \frac{3}{2}\\\\\therefore \ \ \ \displaystyle \frac{a}{x}=\displaystyle \frac{9}{{2n-4}}\\\\\therefore \ \ \displaystyle \frac{6}{{n-1}}=\displaystyle \frac{9}{{2n-4}}\\\\\therefore \ \ 4n-8=3n-3\\\\\therefore \ \ n=5\\\\\therefore \ \ \displaystyle \frac{a}{x}=\displaystyle \frac{6}{{5-1}}\Rightarrow a=\displaystyle \frac{3}{2}x\\\\\therefore 5\left( {\displaystyle \frac{3}{2}x} \right){{x}^{{5-1}}}=240\\\\\therefore {{x}^{5}}=32\Rightarrow x=2\\\\\therefore a=\displaystyle \frac{3}{2}(2)=3\end{array}$

13.       If in the expansion of $ \displaystyle (a + b)^n$, the coefficients of $ \displaystyle p^{\text{th}}$ and $ \displaystyle q^{\text{th}}$ terms are equal, show that $ \displaystyle p+q=n+2$,where $ \displaystyle p\ne q$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {a+b} \right)}^{n}}={{\ }^{n}}{{C}_{r}}{{a}^{{n-r}}}{{b}^{r}}\\\\\therefore \ \ {{p}^{{\text{th}}}}\text{ term}={{\left( {p-1+1} \right)}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{{p-1}}}{{a}^{{n-p+1}}}{{b}^{{p-1}}}\\\\\ \ \ \ {{q}^{{\text{th}}}}\text{ term}={{\left( {q-1+1} \right)}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{{q-1}}}{{a}^{{n-q+1}}}{{b}^{{q-1}}}\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ \text{coefficient of}\ {{p}^{{\text{th}}}}\text{ term}=\text{coefficient of}\ {{q}^{{\text{th}}}}\text{ term}\\\\\ \ \ \ \text{Since}\ p\ne q,\text{ }{}^{n}{{C}_{{p-1}}}\ne {}^{n}{{C}_{{q-1}}}.\\\\\therefore \ \ {}^{n}{{C}_{{p-1}}}={}^{n}{{C}_{{n-q+1}}}\\\\\ \ \ \ p-1=n-q+1\\\\\therefore \ \ p+q=n+2\end{array}$

14.       If the coefficient of $ \displaystyle (m + 1)^{\text{th}}$ term in the expansion of $ \displaystyle (1 + x)^{2n}$ is equal to that of $ \displaystyle (m + 3)^{\text{th}}$ term, then show that $ \displaystyle m = n -1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{2n}}}={{\ }^{{2n}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{\left( {m+1} \right)}^{{\text{th}}}}\text{ term}={}^{{2n}}{{C}_{m}}{{x}^{m}}\\\\\ \ \ \ {{\left( {m+3} \right)}^{{\text{th}}}}\text{ term}={}^{{2n}}{{C}_{{m+2}}}{{x}^{{m+2}}}\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ \ \text{coefficient of }{{\left( {m+1} \right)}^{{\text{th}}}}\text{ term}=\text{coefficient of}\ {{\left( {m+3} \right)}^{{\text{th}}}}\text{ term}\\\\\ \ \ \ {}^{{2n}}{{C}_{m}}={}^{{2n}}{{C}_{{m+2}}}\ (\text{or})\ {}^{{2n}}{{C}_{m}}={}^{{2n}}{{C}_{{2n-m-2}}}\\\\\therefore \ \ \ m=m+2\ \text{which is impossible}\text{.}\ \ (\text{or})\ m=2n-m-2\\\\\therefore \ \ 2m=2n-2\Rightarrow m=n-1\end{array}$

15.       Find the value of $ \displaystyle r$ if the coefficients of $ \displaystyle (2r + 4)^{\text{th}}$ and $ \displaystyle (r - 2)^{\text{th}}$ terms in the expansion of $ \displaystyle (1 + x)^{18}$ are equal.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{18}}}={{\ }^{{18}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{\left( {2r+4} \right)}^{{\text{th}}}}\text{ term}={}^{{18}}{{C}_{{2r+3}}}{{x}^{r}}\\\\\ \ \ \ {{\left( {r-2} \right)}^{{\text{th}}}}\text{ term}={}^{{18}}{{C}_{{r-3}}}{{x}^{r}}\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ \ \text{coefficient of }{{\left( {2r+4} \right)}^{{\text{th}}}}\text{ term}=\text{coefficient of}\ {{\left( {r-2} \right)}^{{\text{th}}}}\text{ term}\\\\\ \ \ \ {}^{{18}}{{C}_{{2r+3}}}={}^{{18}}{{C}_{{r-3}}}\ (\text{or})\ {}^{{18}}{{C}_{{2r+3}}}={}^{{18}}{{C}_{{18-r+3}}}\\\\\therefore \ \ \ 2r+3=r-3\ \ \ (\text{or})\ \ 2r+3=18-r+3\\\\\therefore \ \ \ r=-6\ \text{which is impossible}\text{.}\ \ (\text{or})\ \ 3r=18\\\\\therefore \ \ r=6\end{array}$

16.       Show that the coefficients of $ \displaystyle x^p$ and $ \displaystyle x^q$ in the expansion of $ \displaystyle (1 + x)^{p+q}$ are equal where $ \displaystyle p\ne q$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{p+q}}}={{\ }^{{p+q}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ \text{Coefficient of}\ {{x}^{p}}={{\ }^{{p+q}}}{{C}_{p}}\\\\\ \ \ \ \text{Coefficient of}\ {{x}^{q}}={{\ }^{{p+q}}}{{C}_{q}}\\\\\ \ \ \ \displaystyle \frac{{\text{Coefficient of }{{x}^{p}}}}{{\text{Coefficient of}\ {{x}^{q}}}}=\displaystyle \frac{{^{{p+q}}{{C}_{p}}}}{{^{{p+q}}{{C}_{q}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{^{{p+q}}{{C}_{p}}}}{{^{{p+q}}{{C}_{{p+q-q}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{^{{p+q}}{{C}_{p}}}}{{^{{p+q}}{{C}_{p}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1\\\\\therefore \ \ \text{Coefficient of}\ {{x}^{p}}=\text{Coefficient of}\ {{x}^{q}}\end{array}$

17.       Show that the coefficient of $ \displaystyle x^n$ in the expansion of $ \displaystyle (1 + x)^{2n}$ is twice the coefficient of $ \displaystyle x^n$ in the expansion of $ \displaystyle (1 + x)^{2n-1}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{2n}}}={{\ }^{{2n}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ \text{Coefficient of}\ {{x}^{n}}={{\ }^{{2n}}}{{C}_{n}}\\\\\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{2n-1}}}={{\ }^{{2n-1}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ \text{Coefficient of}\ {{x}^{n}}={{\ }^{{2n-1}}}{{C}_{n}}\\\\\ \therefore \ \ \displaystyle \frac{{^{{2n}}{{C}_{n}}}}{{^{{2n-1}}{{C}_{n}}}}=\displaystyle \frac{{\displaystyle \frac{{2n(2n-1)(2n-2)...(n+3)(n+2)(n+1)}}{{1\times 2\times 3...(n-2)(n-1)n}}}}{{\displaystyle \frac{{(2n-1)(2n-2)...(n+3)(n+2)(n+1)n}}{{1\times 2\times 3...(n-2)(n-1)n}}}}\\\\\therefore \ \ \displaystyle \frac{{^{{2n}}{{C}_{n}}}}{{^{{2n-1}}{{C}_{n}}}}=\displaystyle \frac{{2n(2n-1)(2n-2)...(n+3)(n+2)(n+1)}}{{(2n-1)(2n-2)...(n+3)(n+2)(n+1)n}}\\\\\therefore \ \ \displaystyle \frac{{^{{2n}}{{C}_{n}}}}{{^{{2n-1}}{{C}_{n}}}}=2\Rightarrow {}^{{2n}}{{C}_{n}}=2\cdot {}^{{2n-1}}{{C}_{n}}\end{array}$

18.       For what value of $ \displaystyle m$, the coefficients of $ \displaystyle (2m + 1)^{\text{th}}$ and $ \displaystyle (4m + 5)^{\text{th}}$ terms, in the expansion of $ \displaystyle (1 + x)^{10}$, are equal ?

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{10}}}={{\ }^{{10}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{(2m+1)}^{{\text{th}}}}\ \text{term}={{\ }^{{10}}}{{C}_{{2m}}}{{x}^{{2m}}}\\\\\,\ \ \ {{(4m+5)}^{{\text{th}}}}\ \text{term}={{\ }^{{10}}}{{C}_{{4m+4}}}{{x}^{{4m+4}}}\\\\\ \ \ \ \text{By the problem,}\\\text{ }\\\ \ \ \ \text{Coefficient of}\ {{(2m+1)}^{{\text{th}}}}\ \text{term}=\text{Coefficient of}\ {{(4m+5)}^{{\text{th}}}}\ \text{term}\\\\\ \therefore \ {{\ }^{{10}}}{{C}_{{2m}}}={{\ }^{{10}}}{{C}_{{4m+4}}}\ \ \ (\text{or})\ \ {}^{{10}}{{C}_{{2m}}}={{\ }^{{10}}}{{C}_{{10-4m-4}}}\\\\\therefore \ \ \ 2m=4m+4\ \ (\text{or})\ \ 2m=10-4m-4\\\\\therefore \ \ \ m=-2\ \text{which is impossible}\ \ (\text{or})\ \ 6m=6\\\\\therefore \ \ m=1\end{array}$

19.       If the coefficients of $ \displaystyle (r - 5)^{\text{th}}$ and $ \displaystyle (2r - 1)^{\text{th}}$ terms in the expansion of $ \displaystyle (1 + x)^{34}$ are equal, find $ \displaystyle r$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{{34}}}={{\ }^{{34}}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{(r-5)}^{{\text{th}}}}\ \text{term}={{\ }^{{34}}}{{C}_{{r-6}}}{{x}^{{r-6}}}\\\\\,\ \ \ {{(2r-1)}^{{\text{th}}}}\ \text{term}={{\ }^{{34}}}{{C}_{{2r-2}}}{{x}^{{2r-2}}}\\\\\ \ \ \ \text{By the problem,}\\\text{ }\\\ \ \ \ \text{Coefficient of}\ {{(r-5)}^{{\text{th}}}}\ \text{term}=\text{Coefficient of}\ {{(2r-1)}^{{\text{th}}}}\ \text{term}\\\\\ \therefore \ {{\ }^{{34}}}{{C}_{{r-6}}}={{\ }^{{34}}}{{C}_{{2r-2}}}\ \ \ (\text{or})\ {{\ }^{{34}}}{{C}_{{r-6}}}={{\ }^{{34}}}{{C}_{{34-2r+2}}}\\\\\therefore \ \ \ r-6=2r-2\ \ (\text{or})\ \ r-6=34-2r+2\\\\\therefore \ \ \ r=-4\ \text{which is impossible}\ \ (\text{or})\ \ 3r=42\\\\\therefore \ \ r=14\end{array}$

20.       If the $ \displaystyle 5^{\text{th}}$ term is $ \displaystyle 4$ times the $ \displaystyle 4^{\text{th}}$ term and $ \displaystyle 4^{\text{th}}$ term is $ \displaystyle 6$ times the $ \displaystyle 3^{\text{rd}}$ term in the expansion of $ \displaystyle (1 + x)^n$, find the values of $ \displaystyle n$ and $ \displaystyle x$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {r+1} \right)}^{{\text{th}}}}\text{ term in the expansion }\\\ \ \ \ \text{of}{{\left( {1+x} \right)}^{n}}={{\ }^{n}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{3}^{{\text{rd}}}}\ \text{term}={{\ }^{n}}{{C}_{2}}{{x}^{2}}\\\\\,\ \ \ {{4}^{{\text{th}}}}\ \text{term}={{\ }^{n}}{{C}_{3}}{{x}^{3}}\\\\\ \ \ \ {{5}^{{\text{th}}}}\ \text{term}={{\ }^{n}}{{C}_{4}}{{x}^{4}}\\\\\ \ \ \ \text{By the problem,}\\\text{ }\\\ \ \ {{\ }^{n}}{{C}_{4}}{{x}^{4}}=4{{\ }^{n}}{{C}_{3}}{{x}^{3}}\\\\\therefore \ {{\ }^{n}}{{C}_{3}}\cdot \displaystyle \frac{{n-3}}{4}x=4{{\ }^{n}}{{C}_{3}}\\\\\therefore \ \ \ x=\displaystyle \frac{{16}}{{n-3}}\\\\\ \ \ \ \text{Again,}{{\text{ }}^{n}}{{C}_{3}}{{x}^{3}}=6{{\ }^{n}}{{C}_{2}}{{x}^{2}}\\\\\therefore \ {{\ }^{n}}{{C}_{2}}\cdot \displaystyle \frac{{n-2}}{3}x=6{{\ }^{n}}{{C}_{2}}\\\\\therefore \ \ \ x=\displaystyle \frac{{18}}{{n-2}}\\\\\therefore \ \ \ \displaystyle \frac{{16}}{{n-3}}=\displaystyle \frac{{18}}{{n-2}}\\\\\therefore \ \ \ n=11\\\\\therefore \ \ \ x=\displaystyle \frac{{16}}{{11-3}}=2\end{array}$

21.      If in the expansion of $ \displaystyle (1 + x)^m(1 ֠x)^n$, the coefficients of $ \displaystyle x$ and $ \displaystyle x^2$ are $ \displaystyle 3$ and $ \displaystyle ֶ$, respectively, find $ \displaystyle m$ and $ \displaystyle n$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ {{(1+x)}^{m}}{{(1-x)}^{n}}\\\\\,\ \ \ =\left( {{}^{m}{{C}_{0}}+{}^{m}{{C}_{1}}x+{}^{m}{{C}_{2}}{{x}^{2}}+...} \right)\left( {{}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}-...} \right)\\\\\therefore \ \ \ \ \ \text{Coefficients of}\ x\ \text{in the expansion }\\\ \ \ \ \ \ \text{of}{{(1+x)}^{m}}{{(1-x)}^{n}}\\\\\ \ \ \ \ \ \ ={}^{m}{{C}_{0}}\left( {-{}^{n}{{C}_{1}}} \right)+{}^{m}{{C}_{1}}{}^{n}{{C}_{0}}\\\\\ \ \ \ \ \ \ =m-n\\\\\ \ \ \ \ \ \text{By the problem, }\\\\\ \ \ \ \ \ m-n=3\ \ \ ---(1)\text{ }\\\\\therefore \ \ \ \ \ \text{Coefficients of}\ {{x}^{2}}\ \text{in the expansion }\\\ \ \ \ \ \ \text{of}{{(1+x)}^{m}}{{(1-x)}^{n}}\\\\\ \ \ \ \ \ ={}^{m}{{C}_{0}}\left( {{}^{n}{{C}_{2}}} \right)+{}^{m}{{C}_{2}}{}^{n}{{C}_{0}}+{}^{m}{{C}_{1}}\left( {-{}^{n}{{C}_{1}}} \right)\\\\\ \ \ \ \ =\displaystyle \frac{{n(n-1)}}{2}+\displaystyle \frac{{m(m-1)}}{2}-mn\\\\\ \ \ \ \ \text{By the problem, }\\\\\ \ \ \ \ \displaystyle \frac{{n(n-1)}}{2}+\displaystyle \frac{{m(m-1)}}{2}-mn=-6\\\\\therefore \text{ }\ n(n-1)+m(m-1)-2mn=-12\\\\\ \ \ \ {{n}^{2}}-n+{{m}^{2}}-m-2mn=-12\\\\\ \ \ \ {{m}^{2}}-2mn+{{n}^{2}}-m-n=-12\\\\\ \ \ \ {{(m-n)}^{2}}-(m+n)=-12\\\\\therefore \ \ 9-(m+n)=-12\\\\\therefore \ \ m+n=21\ \ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2m=24\Rightarrow m=12\\\\\ \ \ (2)-(1)\Rightarrow 2n=18\Rightarrow n=9\\\ \end{array}$

22.       Let $ \displaystyle n$ be a positive integer. If the coefficients of $ \displaystyle 2^{\text{nd}},3^{\text{rd}}$ and $ \displaystyle 4^{\text{th}}$ terms in the expansion of $ \displaystyle (1 + x)^n$ are in $ \displaystyle A.P.$, find $ \displaystyle n$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ in the expansion}\\\text{ }\\\ \ \ \ \text{of}\ {{(1+x)}^{n}}={}^{n}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ {{2}^{{\text{nd}}}}\text{ term}={}^{n}{{C}_{1}}x\\\\\ \ \ \ {{3}^{{\text{rd}}}}\text{ term}={}^{n}{{C}_{2}}{{x}^{2}}\\\\\ \ \ \ {{4}^{{\text{th}}}}\text{ term}={}^{n}{{C}_{3}}{{x}^{3}}\\\\\ \ \ \ \text{By the problem,}\\\text{ }\\\ \ \ \ {}^{n}{{C}_{1}},{}^{n}{{C}_{2}},{}^{n}{{C}_{3}}\ \text{is an A}\text{.P}\text{.}\\\\\therefore \ \ {}^{n}{{C}_{2}}-{}^{n}{{C}_{1}}={}^{n}{{C}_{3}}-{}^{n}{{C}_{2}}\\\\\therefore \ \ 2{}^{n}{{C}_{2}}={}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}\\\\\therefore \ \ 2{}^{n}{{C}_{1}}\cdot \displaystyle \frac{{n-1}}{2}={}^{n}{{C}_{1}}+{}^{n}{{C}_{1}}\cdot \displaystyle \frac{{(n-1)(n-2)}}{6}\\\\\therefore \ \ 2{}^{n}{{C}_{1}}\cdot \displaystyle \frac{{n-1}}{2}={}^{n}{{C}_{1}}+{}^{n}{{C}_{1}}\cdot \displaystyle \frac{{(n-1)(n-2)}}{6}\\\\\therefore \ \ 6n-6=1+(n-1)(n-2)\\\\\ \ \ \ \begin{array}{*{20}{l}} {{{n}^{2}}-9n+14=0} \end{array}\,\ \\\\\ \ \ \ (n-2)(n-7)=0\\\\\therefore \ \ n=2\ (\text{or})\ \ n=7\\\\\ \ \ \ \text{Since}\ n\ge 3,\ n=2\ \text{is impossible}\text{.}\\\\\therefore \ \ n=7.\ \ \ \end{array}$