Question

If the middle term of $(1+x{)}^{2n}$ is the greatest term then $x$ lies between

• A. $n-1<x<n$
• B. $\frac{n}{n+1}<x<\frac{n+1}{n}$
• C. $n<x<n+1$
• D. $\frac{n+1}{n}<x<\frac{n}{n+1}$

Answer 1

The correct option is B $\frac{n}{n+1}<x<\frac{n+1}{n}$

The numerically greatest term in $(1+x{)}^n$ will be given by.

$\frac{(1+n)\left(|x|\right)}{1+|x|}$

Now,
In $(1+x{)}^{2n}$
The middle term would be $\frac{2n}{2}+1$ th term

Hence $T_{n+1}$ would be the middle term.

There applying condition for a numerically greatest term we get.

$n<\frac{(2n+1)|x|}{|x|+1}<n+1$ ...(since there would be total of 2n+1 terms).

$=n|x|+n<2n|x|+|x|$ $2n|x|+|x|<n|x|+x+n+1$

$=n<n|x|+|x|$ $n|x|<n+1$

$=n<|x|(n+1)$ $|x|<\frac{n+1}{n}$ ...(i)

$=\frac{n}{n+1}<|x|$ ...(ii)

Hence from i and ii we get

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

Hence answer is Option B