Question

If the coefficients of second, third and fourth term in the expansion of $(1+x{)}^{2n}$ are in A.P, then $2n^2-9n+7$ is equal to

• A. -1
• B. \N
• C. 1
• D. 3/2

Answer 1

The correct option is B \N

$T_2{=}^{2n}{C}_{1}x,T_3{=}^{2n}{C}_2{x}^2,T_4{=}^{2n}{C}_3{x}^3$
Coefficient of $T_1,T_2,T$ are in A.P.
$\Rightarrow {2.}^{2n}{C}_2{=}^{2n}{C}_1{+}^{2n}{C}_3$
$\Rightarrow 2\frac{2n!}{2!(2n-2)!}=\frac{2n!}{(2n-2)!}+\frac{2n!}{3!(2n-3)!}$
$\Rightarrow \frac{2.2n(2n-1)}{2}=2n+\frac{2n(2n-1)(2n-2)}{6}$
$\Rightarrow n(2n-1)=n+\frac{\left(n\right)(2n-1)(2n-2)}{6}$
$\Rightarrow 6(2n^2-n)=6n+4n^3-6n^2+2n$
$\Rightarrow 6n(2n-1)=2n(2n^2-3n+4)$
$\Rightarrow 6n-3=2n^2-3n+4$
$\Rightarrow 0=2n^2-9n+7\Rightarrow 2n^2-9n+7=0$