Question

The sum of coefficients of the two middle terms in the expansion of $(1+x{)}^{2n-1}$ is equal to ${}^{2n-1}{C}_n$. State true or false.

Answer 1

The general term of the expansion $(a+b{)}^n$ is given by

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

Given: $(1+x{)}^{2n-1}$

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

If n is odd, then the middle terms are ${\left(\frac{n+1}{2}\right)}^{th}$and ${\left(\frac{n+3}{2}\right)}^{th}$ term

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

$\frac{(2n-1+3)}{2}=n+1$

So, the middle terms are $n^{th}$ term and $(n+1{)}^{th}$ term

Coefficient of n^th term in the expansion = ${}^{2n-1}{C}_{n-1}$

Coefficient of $(n+1{)}^{th}$ term in the expansion = ${}^{2n-1}{C}_n$

Therefore, the sum of coefficients of the two middle terms

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

=${}^{2n}{C}_n$

$\left[{\because }^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r\right]$

Hence, the given statement is False.