Sum of binomial coefficients with index divisible by 4

We have the followings :

$$(1+1)^n=\binom n0+\binom n1+\binom n2+\binom n3+\binom n4+\binom n5+\cdots$$ $$(1-1)^n=\binom n0-\binom n1+\binom n2-\binom n3+\binom n4-\binom n5+\cdots$$ $$(1+i)^n=\binom n0+\binom n1i-\binom n2-\binom n3i+\binom n4+\binom n5i-\cdots$$ $$(1-i)^n=\binom n0-\binom n1i-\binom n2+\binom n3i+\binom n4-\binom n5i-\cdots$$

Adding these gives $$(1+1)^n+(1-1)^n+(1+i)^n+(1-i)^n=4\left(\binom n0+\binom n4+\binom n8+\cdots\right)$$ I think that you can continue from here.