Question

If $(1-x+x^2{)}^n=a_0+a_{1}x+\dots +a_{2n}{x}^{2n}$, then the value of $a_0+a_2+a_4+\dots +a_{2n}$ is

• A. $3^n+\frac{1}{2}$
• B. $3^n-\frac{1}{2}$
• C. $\frac{3^n-1}{2}$
• D. $\frac{3^n+1}{2}$

Answer 1

The correct option is D $\frac{3^n+1}{2}$

Given, $(1-x+x^2{)}^n=a_0+a_{1}x+\dots +a_{2n}{x}^{2n}\dots (i)$
$x=1$, then from eq. (i)
$1=a_0+a_1+\cdots +a_{2n}\dots \left(ii\right)$
and if $x=-1$, then from eq.(i)
$3^n=a_0-a_1+a_2-a_3+\cdots +a_{2n}\dots \left(iii\right)$
Adding eqs. (ii) and (iii)
$1+3^n=2[a_0+a_2+a_4+\cdots +a_{2n}]$
$\Rightarrow \frac{(1+3^n)}{2}=a_0+a_2+a_4+\cdots +a_{2n}$