\\$Sum the expression, $\\\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+...+\binom{n}{n}\\$by considering the binomial expansions of $ (1+i)^n, (1-i)^n,(1+1)^n $ and $(1-1)^n $ (note that $n$ is a multiple of 4)$\\ i got (by simply adding up all the expansions): \\\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+...+\binom{n}{n}=\frac{1}{2}[\left2^{\frac{n}{2}}\cos\frac{n\pi}{4}+2^{n-1}] can someone do it and confirm (as i don't have answers), thanks :)