Question

If $n$ is a positive integer and ${(1+x+x^2)}^n=\sum _{r=0}^n{a}_r{x}^r$ ,then $(0⩽r⩽2n)$

• A. $a_r=a_{n-r}$
• B. $a_r=a_{2n-r}$
• C. $2a_r=a_{n-r}$
• D. $a_r=2a_{2n-r}$

Answer 1

The correct option is B $a_r=a_{2n-r}$

We have

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

Replace $x$ by $\frac{1}{x}$

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

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

$\Rightarrow$${\sum }_{r=0}^{2n}{a}_r{x}^r$ $=$ ${\sum }_{r=0}^{2n}{a}_r{x}^{2n-r}$

So, equating both the sides, we get $a_r$ $=$ $a_{2n-r}$.