The Binomial Theorem

Observe the following expansion.



(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵



This expression can also be expressed as follows:













In general,if n is a positive integer,





















This expression is known as the Binomial Theorem.



Where the general coefficient or the coefficient of (r + 1)^th term is denoted as follows:

Note:

For the expansion of (x + y)ⁿ,where n is a positive integer,



(i) the binomial coefficients are all integers.

(ii) ⁿC₀ =ⁿC_n =1, ⁿC_n =ⁿC_n-1 = n, ⁿC_r =ⁿC_n-r .

The general term or the (r + 1)^th term of the expansion of (x + y)ⁿ is determined as follows:

(r + 1)^th term = ⁿC_r x^n-r y^r

Example 1

If the coefficients of x⁴ and x⁵ in the expansion of (3 + kx)¹⁰ are equal, find the value k.

(3 + kx)¹⁰ ကိုျဖန္႔္လွ်င္ x⁴ ၏ေျမွွာက္ေဖၚကိန္းသည္ x⁵ ၏ေျမွွာက္ေဖၚကိန္းႏွင့္ တူညီလွ်င္ k ၏တန္ဖိုးကို ရွာပါ။



Solution

(3 + kx)¹⁰

(r + 1)^th term = ¹⁰C_r 3^10-r (kx)^r = ¹⁰C_r 3^10-r k^rx^r



Here ¹⁰C_r 3^10-r k^r is the coefficient of x^r.

coefficient of x⁴ = ¹⁰C₄ 3⁶ k⁴

coefficient ofx⁵ = ¹⁰C₅ 3⁵ k⁵



By the problem,

¹⁰C₄ 3⁶ k⁴ = ¹⁰C₅ 3⁵ k<sup>4

10</sup>C₄ 3⁶ k⁴ = ¹⁰C₄ (6/5) 3⁵ k<sup>5

3 = (6/5)k

k = 5/2</sup>

Example 2

In the expansion of ,a x⁷ is four times the coefficient of x¹⁰. Find the value a.

ကိုျဖန္႔္လွ်င္ x⁷ ၏ေျမွွာက္ေဖၚကိန္းသည္ x¹⁰ ၏ေျမွွာက္ေဖၚကိန္း၏ 4 ဆျဖစ္လွ်င္ a ၏တန္ဖိုးကို ရွာပါ။



Solution







(r + 1)^th term =⁸C_r a^rx^16-3r

To get the coefficient of x⁷, 16-3r = 7 and r =3

To get the coefficient of x¹⁰, 16-3r = 10 and r =2



Therefore,coefficient of x⁷ = ⁸C₃ a³

coefficient of x¹⁰ = ⁸C₂ a²



By the problem,

⁸C₃ a³ = 4 ⁸C₂ a<sup>2

8</sup>C₂ (6/3) a³ = 4 ⁸C₂ a²

a = 2