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 A.P, find $n$.

Answer 1

$(1+2x^2+x^4)(1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3+...)=(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...)(1+{}^n{C}_{1}x+(2+{}^n{C}_2)x^2+({}^n{C}_3+2{}^n{C}_1)x^3+...)=(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...)$

Hence,

$a_1={}^n{C}_1,a_2=2+{}^n{C}_2,a_3={}^n{C}_3+2{}^n{C}_1$

Given, $a_1,a_2,a_3$ are in AP, implies $2a_2=a_1+a_3$

Putting values of $a_1,a_2,a_3$ we get

$2(2+{}^n{C}_2)={}^n{C}_1+{}^n{C}_3+2{}^n{C}_{1}4+2{}^n{C}_2=3{}^n{C}_1+{}^n{C}_3{}^n{C}_3+3{}^n{C}_1-2{}^n{C}_2-4=0\frac{n(n-1)(n-2)}{6}+3n-2\frac{n(n-1)}{2}-4=0n^3-9n^2+26n-24=0n^3-4n^2-5n^2+20n+6n-24=0n^2(n-4)-5n(n-4)+6(n-4)=0(n-4)(n^2-5n+6)=0(n-4)(n^2-3n-2n+6)=0(n-4)(n-3)(n-2)=0n=4,3,2$

$n$ can not be $2$ as ${}^n{C}_3$ wil not be valid,

Hence, $n=4,3$