Question

Let ${(1+x^2)}^2{(1+x)}^n={\sum }_{k=0}^{n+4}{a}_k{x}^k$. If $a_1,a_2$ and $a_3$ are in arithmetic progression, find $n$

Answer 1

Given
${(1+x^2)}^2{(1+x)}^n={\sum }_{k=0}^{n+4}{a}_k{x}^k$
$\Rightarrow (1+{2x}^2+x^4)(1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+....)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....$
Comparing coefficients of $x,x^2,x^3....$, we get
$a_1=n$
$a_2=2+\frac{n(n-1)}{2!}$
$\Rightarrow a_2=\frac{n^2-n+4}{2}$
$a_3=2n+\frac{n(n-1)(n-2)}{3!}$
$\Rightarrow a_3=\frac{n^3-3n^2+14n}{6}$
Since, $a_1,a_2,a_3$ are in A.P.
$\Rightarrow 2a_2=a_1+a_3$
$\Rightarrow n^2-n+4=n+\frac{n^3-3n^2+14n}{6}$
$\Rightarrow n^3-9n^2+26n-24=0$
$\Rightarrow (n-2)(n-3)(n-4)=0$
$\Rightarrow n=2,3,4$