Question

Show that the coefficient of the middle term in the expansion of $(1+x{)}^{2n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x{)}^{2n-1}$

Answer 1

Step $1$: Middle terms of given binomial expansion
For $(1+x{)}^{2n}$
Since $2n$ is even,
middle term $={(\frac{2n}{2}+1)}^{th}$term
$=(n+1{)}^{th}$ term

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

Hence, the coefficient of middle term is ${}^{2n}{C}_n$
Now, middle term of $(1+x{)}^{2n-1}$
Since $2n-1$ is odd.

There will be two middle terms $={\left(\frac{(2n-1)+1}{2}\right)}^{th}$ and ${(\frac{(2n-1)+1}{2}+1)}^{th}$term

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

$=T_n$ and $T_{n+1}$

For $T_n$ in $(1+x{)}^{2n-1}$

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

$\Rightarrow T_n={}^{2n-1}{C}_{n-1}(x{)}^{n-1}$
Hence, coefficient is ${}^{2n-1}{C}_{n-1}$

Similarly,
For $T_{n+1}$ in $(1+x{)}^{2n-1}$

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

$T_{n+1}={}^{2n-1}{C}_n{x}^n$

Hence, coefficient is ${}^{2n-1}{C}_n$

Step $2$ : Solve for prove
Now we have to prove coefficient of middle term of $(1+x{)}^{2n}$ = sum of coefficient of middle terms of $(1+x{)}^{2n-1}$

${}^{2n}{C}_n{=}^{2n-1}{C}_{n-1}{+}^{2n-1}{C}_n$

L.H.S ${=}^{2n}{C}_n$
$=\frac{\left(2n\right)!}{n!(2n-n)!}$

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

R.H.S.${=}^{2n-1}{C}_{n-1}{+}^{2n-1}{C}_n$

$=\frac{(2n-1)!}{(n-1)![2n-1-n+1]!}+\frac{(2n-1)!}{n![2n-1-n]!}$

$=\frac{(2n-1)!}{(n-1)!n!}+\frac{(2n-1)!}{n!(n-1)!}$

$=2\times \frac{(2n-1)!}{(n-1)!n!}\times \frac{n}{n}$

$=\frac{2n(2n-1)!}{n!\left[n\right(n-1)!]}=\frac{\left(2n\right)!}{n!n!}$

Hence L.H.S. $=$ R.H.S
Hence proved.