Question

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

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

Answer 1

The correct option is C $512$

${(1+x+x^2+x^3)}^5=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots +a_{15}{x}^{15}$
Putting $x=1$ and $x=-1$ alternatively, we have
$$\begin{array}{c}{a}_0+a_1+a_2+a_3+\cdots +a_{15}=4^5\\ a_0-a_1+a_2-a_3+\cdots -a_{15}=0\end{array}$$
Adding $\left(1\right)$ and $\left(2\right)$, we have $2(a_0+a_2+a_4+\cdots +a_{14})=4^5$
$\Rightarrow a_0+a_2+a_4+\cdots +a_{14}=2^9=512$

Alternate Solution:
$(1+x+x^2+x^3{)}^5=(1+x{)}^5(1+x^2{)}^5$
Now $(1+x^2{)}^5$ have only even powers of $x$
so, the sum will be $2^5$
and sum of even power coefficient in $(1+x{)}^5$ will be $2^{5-1}=2^4$
Hence required sum is
$2^4\times 2^5=2^9=512$