Question

The sum of the coefficients of even powers of $x$ in the expansion

$(1-x+x^2-x^3{)}^5$ is

• A. $512$
• B. $0$
• C. $-512$
• D. $510$

Answer 1

The correct option is A $512$

The sum of coefficients of even powers of $x$ is an even Function.
Hence $\frac{f\left(1\right)+f(-1)}{2}=$sum of coefficients of even powers of $x$.
Here $f\left(x\right)=(1-x+x^2-x^3{)}^5$.
Therefore
$f\left(1\right)=0$ ...(i)
$f(-1)=4^5=2^{10}$ ...(ii)
Hence $\frac{f\left(1\right)+f(-1)}{2}$
$=\frac{2^{10}}{2}$
$=2^9$
$=512$