A probability problem by Rohit Sharma

Putting ${x=i}$ , where ${i}=\displaystyle\sqrt{-1}$

$(1+{i}+{i}^2+{i}^3)^7={a_0}+{a_1}{i}+{a_2}{i}^2+...+{a_{21}}{i}^{21}$

$0=({a_0}-{a_2}+{a_4}-...)+{i}({a_1}-{a_3}+{a_5}...)$

Therefore ,
$({a_0}-{a_2}+{a_4}-...)=0$ ....(1)

Putting ${x=1,-1}$ , we get
$4^7=({a_0}+{a_1}+{a_2}+...)$ and $0=({a_0}-{a_1}+{a_2}-...)$ respectively.

By adding above two equations we get ,

$4^7=2({a_0}+{a_2}+{a_4}+...)$
or $2^{13}=({a_0}+{a_2}+{a_4}+...)$ ....(2)

Now adding (1) and (2) we get ,

$2({a_0}+{a_4}+{a_8}+{a_{12}}+{a_{16}}+{a_{20}})=2^{13}$

Hence , ${a_0}+{a_4}+{a_8}+{a_{12}}+{a_{16}}+{a_{20}}= 4096$