Question

If ${(1+2x+x^2)}^5=\sum _{k=0}^{15}{a}_k{x}^k$then $\sum _{k=0}^7{a}_{2k}=$

Answer 1

Finding the value of $\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{7}}{\mathit{a}}_{\mathbf{2}\mathbf{k}}\mathbf{}\mathbf{:}$

Given that ${(1+2x+x^2)}^5=\sum _{k=0}^{15}{a}_k{x}^k$

${(1+x)}^{10}=\sum _{k=0}^{15}{a}_k{x}^k$

$10C_0+10C_{1}x+10C_2{x}^2+....+10C_{10}{x}^{10}=a_0+a_{1}x+a_2{x}^2+...+a_{10}{x}^{10}+..+a_{15}{x}^{15}$

Comparing the coefficients of odd terms

$a_0=10C_0,a_2=10C_2,......a_{10}=10C_{10},....a_{12}=a_{14}=0$

$\sum _{k=0}^7{a}_{2k}=a_0+a_2+a_4+....a_{14}$

$$\begin{gathered} =1+10C_2+10C_4+10C_6+10C_8+10C_{10}+0+0 \\ =2^{10-1} \\ =2^9 \end{gathered}$$

Hence, the value of $\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{7}}{\mathit{a}}_{\mathbf{2}\mathbf{k}}\mathbf{=}$ is ${\mathbf{2}}^{\mathbf{9}}$.