Question

Show that if the greatest term in the expansion of ${(1+x)}^{2n}$ has also the greatest coefficient, then $x$ lies between $\frac{n}{n+1}$ and $\frac{n+1}{n}$

Answer 1

In the expansion of $(1+x{)}^{2n}$, middle term is $(2n/2+1{)}^{th}$ i.e., $(n+1{)}^{th}$ term, we know that from binomial expansion middle term has greatest coefficient.

$\therefore T_n<T_{n+1}>T_{n+2}$

$\Rightarrow \frac{T_{n+1}}{T_n}=\frac{{}^{2n}{C}_n{x}^n}{{}^{2n}{C}_{n-1}{x}^{n-1}}=\frac{2n-n+1}{n}x$

$\Rightarrow \frac{T_{n+1}}{T_n}>1$ or $\frac{n+1}{n}\cdot x>1$

(or) $x>\frac{n}{n+1}$ …………$\left(1\right)$

and $\frac{T_{n+2}}{T_{n+1}}=\frac{{}^{2n}{C}_{n+1}{x}^{n+1}}{{}^{2n}{C}_n{x}^n}=\frac{2n-(n+1)+1}{n+1}x$

$=\frac{n}{n+1}x$

$\Rightarrow \frac{T_{n+1}}{T_{n+1}}<1$

$\Rightarrow \frac{n}{n+1}x<1$

$\Rightarrow x<\frac{n+1}{n}$ ………..$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$, we get

$\frac{n}{n+1}<x<\frac{n+1}{n}$.