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 A.P,Find value of n

Answer 1

$(1+x^2{)}^2(1+x{)}^n$

$=(1+2x^2+x^4)(1+x)=(1+2x^2+x^4){(}^n{C}_{0}x{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_2{x}^3+......{+}^n{C}_n{x}^n)$

${=}^n{C}_0{+}^{2n}{C}_0{x}^2{+}^n{C}_0{x}^4$

${=}^n{C}_{1}x{+}^{2n}{C}_1{x}^3{+}^n{C}_1{x}^5$

${=}^n{C}_2{x}^2+2n{C}_2{x}^4{+}^n{C}_2{x}^6$

$\begin{array}{c}.\\ .\\ .\end{array}$

$=\sum _{k-0}^{n+4}{a}_k{x}_k$

$a_1{=}^n{C}_1$

$a_2{=}^{2n}{C}_0{+}^n{C}_2$

$a_3{=}^{2n}{C}_1$

$\frac{{}^n{C}_1{+}^{2n}{C}_1}{2}{=}^{2n}{C}_0{+}^n{C}_2$ $(\because a_1,a_2,a_3$ are in $AP$)

$\frac{3n}{2}=2+\frac{n(n-1)}{2}$

$\therefore 3n=4+n^2-n$

$n^2-4n+4=0$

$\therefore (n-2{)}^2=0$

$\therefore n=2$