Question

The greatest coefficient in the expansion of ${\left(1+x\right)}^{2n+1}$ is.

• A. $\frac{\left(2n+1\right)!}{n!\left(n+1\right)!}$
• B. $\frac{\left(2n+1\right)!}{6!\left(n+1\right)!}$
• C. $\frac{\left(2n+1\right)!}{{\left[\left(n+1\right)!\right]}^2}$
• D. $\frac{\left(2n\right)!}{{\left(n!\right)}^2}$

Answer 1

The correct option is A $\frac{\left(2n+1\right)!}{n!\left(n+1\right)!}$

Given expression is $(1+x{)}^{2n+1}$

The greatest coefficient in the expansion of (1+x)^n is coefficient of middle

terms here $2n+1$ is odd

$\therefore$ the middle terms are ${\left(\frac{2n+2}{2}\right)}^{th},{\left(\frac{2n+4}{2}\right)}^{th}$ i.e. $(n+1{)}^{th}$ and $(n+2{)}^{th}$

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

The greatest coefficient in the expansion of $(1+x{)}^{2n+1}$ are

${}^{2n+1}{C}_n,{}^{2n+1}{C}_{n+1}$

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

${}^{2n+1}{C}_n=\frac{(2n+1)!}{n!(n+1)!}$