Question

Let $n$ be a positive integer such that ${(1+x+x^2)}^n=a_0+a_{1}x+a_2{x}^2+...+a_{2n}{x}^{2n}$ then $a_r=$

• A. $a_{n-r},0\le r\le 2n$
• B. $a_{2n},0\le r\le 2n$
• C. $a_{2n-r},0\le r\le 2n$
• D. none of these

Answer 1

The correct option is C $a_{2n-r},0\le r\le 2n$

We have ${(1+x+x^2)}^n=\sum _{r=0}^{2n}{a}_r{x}^r$............(1)

Replacing $x$ by $\frac{1}{x}$ we get

${(1+\frac{1}{x}+\frac{1}{x^2})}^n=\sum _{r=0}^{2n}{a}_r\frac{1}{x^r}$

Multiplying both sides by $x^{2n}$ we get,

${(1+x+x^2)}^n=\sum _{r=0}^{2n}{a}_r{x}^{2n-r}$...............(2)

From (1) and (2) we have, $\sum _{r=0}^{2n}{a}_r{x}^r=\sum _{r=0}^{2n}{a}_r{x}^{2n-r}$

Equating the coefficients of $x^{2n-r}$, we have

$a_{2n-r}=a_r$ for $0\le r\le 2n$