Question

If the sum of the series ${}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}$ is $2^a(2^a+1)$, then the value of $a$ is

Answer 1

Given, ${}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}=2^{38}+2^a$

$(1+x{)}^{40}+(1-x{)}^{40}=2[{}^{40}{C}_0+{}^{40}{C}_2{x}^2+{}^{40}{C}_4{x}^4+\cdots +{}^{40}{C}_{40}{x}^{40}]...\left(1\right)$
Put $x=1$ in equation $\left(1\right)$
$2^{39}={}^{40}{C}_0+{}^{40}{C}_2+{}^{40}{C}_4+\cdots +{}^{40}{C}_{40}...\left(2\right)$

Put $x=i(i=\sqrt{-1})$ in equation $\left(2\right)$
$\frac{1}{2}\left[\right(1+i{)}^{40}+(1-i{)}^{40}]={}^{40}{C}_0-{}^{40}{C}_2+{}^{40}{C}_4-\cdots +{}^{40}{C}_{40}...\left(3\right)$

Add equation $\left(2\right)$ and $\left(3\right),$ we get
$2^{39}+\frac{1}{2}\times (\sqrt{2}{)}^{40}[{\left(e^{i\pi /4}\right)}^{40}+{\left(e^{-i\pi /4}\right)}^{40}]=2[{}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}]$
$\Rightarrow 2^{39}+2^{19}\left(2\right)=2[{}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}]$
$\Rightarrow 2^{38}+2^{19}={}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}$
$\Rightarrow 2^{19}(2^{19}+1)={}^{40}{C}_0+{}^{40}{C}_4+{}^{40}{C}_8+\cdots +{}^{40}{C}_{40}$