If $\left(1+x+x^2+x^3\right)^5=\sum_{k=0}^{15} a_k x^k$, then $\sum_{k=0}^7 a_{2 k}$ is equal to

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Given, }\left(1+x+x^2+x^3ight)^5=\sum_{k=0}^{15} a_k x^k \ & \Rightarrow[(1+x)+x(1+x)]^5=\sum_{k=0}^{15} a_k x^k \ & \Rightarrow(1+x)^{10}=a_0 x^0+a_1 x+a_2 x^2+\ldots+a_{15} x^{15} \ & \Rightarrow \quad{ }^{10} C_0+{ }^{10} C_1 x+{ }^{10} C_2 x^2+\ldots+{ }^{10} C_{10} x^{10} \ & \quad=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots+a_{15} x^{15} \end{aligned} $ On equating the coefficient of constant and even powers of $x$, we get $ \begin{aligned} a_0 & ={ }^{10} C_0, a_2={ }^{10} C_2, \ a_4 & ={ }^{10} C_4, \ldots, a_{10}={ }^{10} C_{10}, \ a_{12} & =a_{14}=0 \ 	herefore \quad \sum_{k=0}^7 a_2 k & ={ }^{10} C_0+{ }^{10} C_2+{ }^{10} C_4+{ }^{10} C_6 \ & +{ }^{10} C_8+{ }^{10} C_{10}+0+0 \ & =2^{10-1}=2^9 \ & =512 \end{aligned} $

If $\left(1+x+x^2+x^3\right)^5=\sum_{k=0}^{15} a_k x^k$, then $\sum_{k=0}^7 a_{2 k}$ is equal to

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