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}$ |
