Question

prove that the middle term in the expansion of
$\huge{ \bf {\red {\boxed{(1 + x) ^{2n} }}}} \:$
is

$\huge{\bf \frac{1.3.5.....(2n - 1)}{n \: ! } {2}^{n}. {x}^{n}. }$
​

Answer 1

Hey!

As middle term of an even exponent = index /2 + 1

= 2n/2 + 1

= n+1th term.

Refer attachment!

Answer 2

Step-by-step explanation:

Given: (1 + x)²ⁿ

The index(2n) is even.

$\therefore Middle \ term = T_{\frac{2n+2}{2}} = T_{n+1} = 2nC_{n}x^n$

$\therefore \ coefficient \ of \ middle term \ in \ (1+x)^{2n} = 2nC_{n}$

$\Longrightarrow T_{n+1} = 2nC_{n} x^n$

$\Longrightarrow \frac{2n!}{n!n!} * x^n$

$\Longrightarrow \frac{1.2.3.4.5..(2n-1)(2n)}{(n!n!)}* x^n$

$\Longrightarrow \frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{(1.2.3...n).n!} * x^n$

$\Longrightarrow \frac{[1.3.5...(2n-1)].2^n[1.2.3...n]}{(1.2.3...n).n!} * x^n$

$\Longrightarrow \frac{[1.3.5..(2n-1)] * 2^n}{n!} * x^n$

$\Longrightarrow {\boxed{\frac{[1.3.5..(2n-1)]}{n!} \ 2^n * x^n}}$

Hope it helps1