Question

The sum of the coefficients of middle terms in the expansion of $(1+x{)}^{2n-1}$

• A. $\left(2n\right)!$
• B. $\frac{\left(2n\right)!}{n!}$
• C. $\frac{\left(2n\right)!}{(n!{)}^2}$
• D. $\frac{(2n-1)!}{n!}$

Answer 1

The correct option is C $\frac{\left(2n\right)!}{(n!{)}^2}$

In the following binomial expansion there would total of
$2n-1+1$ terms
$=2n$ number of terms.
Since the total number of terms would be even, there would be two middle terms.
$\frac{2n-1+1}{2}$th term and $\frac{2n-1+1}{2}+1$th term.
Hence $n^{th}$ and $(n+1{)}^{th}$ terms.
Coefficient of $n^{th}$ term
$=T_{(n-1)+1}$
$={}^{2n-1}{C}_{n-1}$ ...(i)
Similarly, coefficient of $(n+1{)}^{th}$ term
$=T_{\left(n\right)+1}$
$={}^{2n-1}{C}_n$ ...(i)
Adding (i) and (ii), we get
${}^{2n-1}{C}_{n-1}+{}^{2n-1}{C}_n$
${=}^{2n}{C}_n$ ...(from identities of binomial coefficients)
$=\frac{2n!}{n!n!}$
$=\frac{2n!}{(n!{)}^2}$