Complex in Binomial

$\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ r }\left( \begin{matrix} n \\ r \end{matrix} \right) \left[ { i }^{ r }+{ i }^{ 2r }+{ i }^{ 3r }+{ i }^{ 4r } \right] }$ can be written as

$\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ r }\left( \begin{matrix} n \\ r \end{matrix} \right) \left[ { i }^{ r }+{ \left( -1 \right) }^{ r }+{ \left( -i \right) }^{ r }+1 \right] }$

On more simplification we get

$\sum _{ r }^{ n }{ \left[ \left( \begin{matrix} n \\ r \end{matrix} \right) { \left( -i \right) }^{ r }+\left( \begin{matrix} n \\ r \end{matrix} \right) +\left( \begin{matrix} n \\ r \end{matrix} \right) { i }^{ r }+\left( \begin{matrix} n \\ r \end{matrix} \right) { \left( -1 \right) }^{ r } \right] }$

thus the above can be represented as : ${ \left( 1-i \right) }^{ n }+{ \left( 1+1 \right) }^{ n }+{ \left( 1+i \right) }^{ n }+{ \left( 1-1 \right) }^{ n }$ . which can be written in a more simplified way by using euler's formula i.e.

${ \left( \sqrt { 2 } \right) }^{ n }{ \left( \cos { \frac { \pi }{ 4 } } +i\sin { \frac { \pi }{ 4 } } \right) }^{ n }+{ \left( \sqrt { 2 } \right) }^{ n }{ \left( \cos { \frac { \pi }{ 4 } } -i\sin { \frac { \pi }{ 4 } } \right) }^{ n }+{ 2 }^{ n }$

On further simplification we get

${ 2 }^{ n }+{ 2 }^{ \left( n/2+1 \right) n }\left( \cos { \frac { n\pi }{ 4 } } \right)$

therefore by comparing we get

A=2, B=1, C=2, D=1, E=2, F=4 thus A+B+C+D+E+F = 12