Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Statement is true, so option $A$ is correct.

Proof :- Middle term of $(1+X{)}^{2n}$

Since $2n$ is even middle term = ${(\frac{2n}{2}+1)}^{th}=(n+1{)}^{th}$ term = $T_{n+1}$

we know that general & term of expansion $(a+b{)}^n$ is -

$T_{n+1}={}^n{C}_r{a}^{n-r}{b}^r$

here, $r=n,n=2n,a=1,b=x$

putting values

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

$T_{nH}={}^{2n}{C}_n(x{)}^n......(A)$

Here, coefficient is ${}^{2n}{C}_n$

$\Rightarrow$ middle term of $(1+x{)}^{2n-1}$

Since, $(2n-1)$ is add

There will be 2 middle terms

$\frac{(2n-1)+1}{2}$ and $(\frac{(2n-1)+1}{2}+1)$ term

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

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

we know that general term for expansion $(a+b{)}^n$

$T_{rH}={}^n{C}_r{a}^{n-r}{b}^r$

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

Putting $r=n-1,n=2n-1,a=1\&b=x$

$T_n={}^{2n-1}{C}_{n-1}(1{)}^{2n-1-(n-1)}{x}_{(n-1)}={}^{2n-1}{C}_{n-1}{x}^{(n-1)}$

Coefficient pf middle term $\left(n^{th}term\right)={}^{2n-1}{C}_{n-1}$

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

Sum of the coefficent of the middle term in the expansion of $(1+X{)}^{2n-1}$

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

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

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

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

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

Equation $\left(A\right)$ = Equation $\left(B\right)$