Question

If ${(1+x+x^2+x^3)}^5=\sum _{k=0}^{15}{a}_k{x}^k$ then $\sum _{k=0}^7{a}_{2k}$ is equal to

• A. $128$
• B. $256$
• C. $512$
• D. $1024$

Answer 1

Answer: (C)

$512$

Given, ${(1+x+x^2+x^3)}^5=\sum _{k=0}^{15}{a}_k{x}^k$
$\Rightarrow {[(1+x)+x(1+x)]}^5=\sum _{k=0}^{15}{a}_k{x}^k$
$\Rightarrow {(1+x)}^{10}=a_0{x}^0+a_{1}x+a_2{x}^2+\cdots +a_{15}{x}^{15}$
${\Rightarrow }^{10}{C}_0{+}^{10}{C}_{1}x{+}^{10}{C}_2{x}^2+\cdots {+}^{10}{C}_{10}{x}^{10}$
$=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots +a_{15}{x}^{15}$
On equating the coefficient of constant and even powers of $x$, we get
$a_0{=}^{10}{C}_0,a_2{=}^{10}{C}_2,$
$a_4{=}^{10}{C}_4,\dots ,a_{10}{=}^{10}{C}_{10},$
$a_{12}=a_{14}=0$
$\therefore \sum _{k=0}^7{a}_{2}k{=}^{10}{C}_0{+}^{10}{C}_2{+}^{10}{C}_4{+}^{10}{C}_6{+}^{10}{C}_8{+}^{10}{C}_{10}+0+0$
$=2^{10-1}=29$
$=512$