Question

The coefficient of middle term in the expansion of $(1+x{)}^{40}$ is

• A. $\frac{\mathrm{1.3.5}\cdots 39}{20!}{.2}^{20}$
• B. $\frac{\mathrm{1.3.5}\cdots 39}{20!}$
• C. $\frac{40}{20}!$
• D. $40!$ ${20}^{20}$

Answer 1

The correct option is A $\frac{\mathrm{1.3.5}\cdots 39}{20!}{.2}^{20}$

Middle term $=\frac{N}{2}+1$
$=\frac{2n}{2}+1$
$=(n+1)thterm$
Now consider the following
$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$...(i)
Where $T$ represents the term
Here the middle term is the $(n+1{)}^{th}$ term.
Hence, $r+1=n+1$
$r=n$
Substituting in (i), we get
${}^{2n}{C}_n{1}^{2n-n}{x}^n$
$={}^{2n}{C}_n{x}^n$
Therefore the coefficient of the middle term is ${}^{2n}{C}_n$
Substituting $n=20$, we get
${}^{40}{C}_{20}$
$=\frac{40!}{20!20!}$
Dividing by $20!$ and separating the multiples of two in the powers of two we get,
Middle term $=\frac{\left(\mathrm{1.3.5.7....39}\right)2^{20}}{20!}$