Question

Let $(1+x^2{)}^2(1+x{)}^n=\sum _{k=0}^{n+4}{a}_k{x}^k$. If $a_1,a_2,a_3$ are in arithmetic progression, then the sum of all the possible values of n is

Answer 1

$\text{L.H.S.}=(1+2x^2+x^4){(}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3+\cdots )\text{R.H.S.}=a_0+a_{1}x+a_2{x}^2+\cdots \therefore a_1{=}^n{C}_1,a_2{=}^n{C}_2+2{\cdot }^n{C}_0\text{and}{a}_3{=}^n{C}_3+2{\cdot }^n{C}_1\because a_1,a_2\text{and}{a}_3\text{are in arithmetic progression,}\therefore 2a_2=a_1+a_3\Rightarrow 2(\frac{n(n-1)}{2}+2)=n+\frac{n(n-1)(n-2)}{6}+2n\text{The above equation is true for}n=2,3,4\therefore \text{Required Sum}=2+3+4=9.$