Question

Shew that the middle term in the expansion of $(1+x{)}^{2n}$ is $\frac{\mathrm{1.3.5...}(2n-1)}{\left(n\right)!}{2}^n{x}^n$.

Answer 1

Number of terms $2n$ which is even

So,

Middle term $={(\frac{2n}{2}+1)}^{th}$ term

$=(n+1{)}^{th}$ term

Hence, we need to find $T_{n+1}$

For $T_{n+1},$

Putting $n=2n,r=n,a=1$ and $b=x$

$T_{n+1}={}^{2n}{C}_n\left(1{)}^{2n-n}\right(x{)}^n$

$=\frac{\left(2n\right)!}{n!(2n-n)!}.(1{)}^n.x^n$

$=\frac{\left(2n\right)!}{n!n!}.x^n$

$=\frac{2n(2n-1)(2n-2)....4\times 3\times 2\times 1}{n!n!}{x}^n$

$=\frac{\left[\right(2n-1\left)\right(2n-3)...\times 5\times 3\times 1]\left[\right(2n\left)\right(2n-2)...\times 4\times 2]}{n!n!}{x}^n$

$=\frac{[1\times 3\times 5\times ...\times (2n-3\left)\right(2n-1\left)\right][2\times 4\times \mathrm{6...}\times (2n-2)\times 2n]}{n!n!}{x}^n$

$=\frac{[1\times 3\times \mathrm{5...}\times (2n-3\left)\right(2n-1\left)\right]\left[\right(2\times 1)\times (2\times 2)\times (2\times 3)\times ...\times 2(n-1)\times 2n]}{n!n!}{x}^n$

$=\frac{[1\times 3\times \mathrm{5....}\times (2n-3\left)\right(2n-1\left)\right(2\times 2\times 2\times \mathrm{2...}\times 2\left)\right[1\times 2\times \mathrm{3...}(n-1)n}{n!n!}{x}^n$

$=\frac{[1\times 3\times \mathrm{5...}\times (2n-3\left)\right(2n-1\left)\right]2^n[1\times 2\times \mathrm{3...}(n-1\left)n\right]}{n!n!}{x}^n$

$=\frac{[1\times 3\times \mathrm{5...}\times (2n-3\left)\right(2n-1\left)\right]2^n(n!)}{n!n!}{x}^n$

$=\frac{1\times 3\times \mathrm{5...}\times (2n-1)}{n!}{2}^n.x^n$