Compute the sum $$\binom{n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots$$
I use the $(1+i)^n$ and $(1+1)^n$ and I obtained: $$(1+i)^n + (1+1)^n=\binom{n}{0}+\binom{n}{1}i+\binom{n}{2}(-1)+\binom{n}{3}(-i)+\binom{n}{4}+\cdots $$$$+\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+ \cdots$$ After reductions: $$(1+i)^n + (1+1)^n=2\left(\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots\right)+2\left(\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots+ \binom{n}{\frac{n}{2}-1}\right)$$ $$(1+i)^n + (1+1)^n=2\left(\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots\right)+2^{\frac{n}{2}-1}$$
Then: $$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots=\frac{(1+i)^n + (1+1)^n-2^{\frac{n}{2}-1}}{2}$$
I'm not very sure that the result is correct, it seems to me that something is missing or I approached wrong this sum.
