Question

The sum of the coefficients of even power of $x$ in the expansion of ${(1+x+x^2+x^3)}^5$ is

• A. $64$
• B. $512$
• C. $128$
• D. $256$

Answer 1

The correct option is B $512$

${(1+x+x^2+x^3)}^5=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots +a_{15}{x}^{15}$

Putting $x=1$ and $x=-1$ alternatively, we get
$a_0+a_1+a_2+a_3+\cdots +a_{15}=4^5\cdots \left(1\right)a_0-a_1+a_2-a_3+\cdots -a_{15}=0\cdots \left(2\right)$
Adding equation $\left(1\right)$ and $\left(2\right)$, we have $2(a_0+a_2+a_4+\cdots +a_{14})=4^5\Rightarrow a_0+a_2+a_4+\cdots +a_{14}=2^9\therefore a_0+a_2+a_4+\cdots +a_{14}=512$