Question

If $n$ is even positive integer, then the condition that the greatest term in the expansion of $(1+x{)}^n$ may also have the greatest coefficient is

• A. $\frac{n}{n+2}<x<\frac{n+2}{n}$
• B. $\frac{n}{n+1}<x<\frac{n+1}{n}$
• C. $\frac{n+1}{n+2}<x<\frac{n+2}{n+1}$
• D. $\frac{n+2}{n+3}<x<\frac{n+3}{n+2}$

Answer 1

The correct option is A $\frac{n}{n+2}<x<\frac{n+2}{n}$

$n$ is even so the greatest coefficient term will be
${(\frac{n}{2}+1)}^{th}$
As this term is greatest so,
${}^n{C}_{n/2}{x}^{n/2}>{}^n{C}_{\left(n/2\right)+1}{x}^{\left(n/2\right)+1}$
And,
${}^n{C}_{n/2}{x}^{n/2}>{}^n{C}_{\left(n/2\right)-1}{x}^{\left(n/2\right)-1}$
$\frac{{}^n{C}_{n/2}}{{}^n{C}_{\left(n/2\right)+1}}>x\Rightarrow \frac{(\left(\frac{n}{2}\right)-1)!(\left(\frac{n}{2}\right)+1)!}{\left(n/2\right)!\left(n/2\right)!}>x\Rightarrow \frac{n+2}{n}>x$

${}^n{C}_{n/2}{x}^{n/2}>{}^n{C}_{\left(n/2\right)-1}{x}^{\left(n/2\right)-1}$
$\frac{{}^n{C}_{n/2}}{{}^n{C}_{\left(n/2\right)-1}}x>1\Rightarrow \frac{n+2}{n}x>1\Rightarrow x>\frac{n}{n+2}$

So the required condition is
$\frac{n}{n+2}<x<\frac{n+2}{n}$