Question

If ${(1+x^2)}^2{(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\cdot \cdot \cdot ,$ and if $C_0,C_1,C_2$ are in A.P., find $n$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A), (B)

3
2

Given,
$(1+x^2{)}^2(1+x{)}^n=C_0+C_{1}x+C_2{x}^2...$ and $C_0,C_1,C_2$ are in AP
$(1+x^4+2x^2){(}^n{\text{C}}_0$+${}^n{\text{C}}_{1}x{+}^n{\text{C}}_2{x}^2...)=C_0+C_{1}x+C_2{x}^2\dots$
Comparing the terms we get,
${}^n{\text{C}}_0=C_0$
${}^n{\text{C}}_1=C_1$
${}^n{\text{C}}_2+2{}^n{\text{C}}_0=C_2$
We know,
$C_0+C_2=2C_1$ (Since they are in A.P.)
$3\times {}^n{\text{C}}_0{+}^n{\text{C}}_2=2{\times }^n{\text{C}}_1$
$\frac{3.n!}{n!}+\frac{n!}{(n-2)!2!}=2\times \frac{n!}{(n-1)!}$
$3+\frac{n(n-1)}{2}=2n$
$6+n^2-n=4n$
$n^2-5n+6=0$
$(n-2)(n-3)=0$
Therefore,
$n=2$ or $3$