Question

Let $(1+x^2{)}^2(1+x{)}^n=\sum _{k=0}^{n+4}{a}_k{x}^k.$
The $a_1,a_2\text{and}{a}_3$ are in A.P, then the possible values of $n$ is/are

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $2$
C $3$
D $4$

L.H.S
$=(1+2x^2+x^4)(1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3+...+{}^n{C}_n{x}^n)$

R.H.S
$=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+.....a_{n+4}{x}^{n+4}$
So by comparing the coefficients, we can write
$a_1={}^n{C}_1$
$a_2={}^n{C}_2+2$
$a_3={}^n{C}_3+2\left({}^n{C}_1\right)$

We know that $a_1,a_2,a_3$ are in A.P so,
$\Rightarrow 2a_2=a_1+a_3$
$\Rightarrow 2({}^n{C}_2+2)={}^n{C}_1+{}^n{C}_3+2{}^n{C}_1$
$\Rightarrow n(n-1)+4=3n+\frac{n(n-1)(n-2)}{6}$
$\Rightarrow n^3-9n^2+26n-24=0$
$\Rightarrow (n-2)(n^2-7n+12)=0$
$\Rightarrow (n-2)(n-3)(n-4)=0$
$\Rightarrow n=2,3,4$