### A result using complex numbers

Suppose we expand out the expressions $$(1+1)^n$$, $$(1+i)^n$$, $$(1-1)^n$$, $$(1-i)^n$$ using the binomial theorem. This gives (for $$n>0$$),

\begin{align*} \dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + \dbinom{n}{3} + \dots + \dbinom{n}{n} &= 2^n\\ \dbinom{n}{0} + i\dbinom{n}{1} - \dbinom{n}{2} - i\dbinom{n}{3} + \dots + i^n\dbinom{n}{n} &= (1+i)^n\\ \dbinom{n}{0} - \dbinom{n}{1} + \dbinom{n}{2} - \dbinom{n}{3} + \dots + (-1)^n\dbinom{n}{n} &= 0^n\\ \dbinom{n}{0} - i\dbinom{n}{1} - \dbinom{n}{2} + i\dbinom{n}{3} + \dots + (-i)^n\dbinom{n}{n} &= (1-i)^n. \end{align*}

Now add these equations and divide by 4 to get

$\dbinom{n}{0} + \dbinom{n}{4} + \dbinom{n}{8} + \dots = \dfrac{1}{4}\Big(2^n + (1-i)^n + (1+i)^n\Big).$

It is easy to show, using the polar form, that

$(1-i)^n + (1+i)^n = 2(\sqrt{2})^n \cos \big(\dfrac{n\pi}{4}\big).$

Thus, we have

$\dbinom{n}{0} + \dbinom{n}{4} + \dbinom{n}{8} + \dots = \dfrac{1}{4}\Big(2^n + 2(\sqrt{2})^n \cos \big(\dfrac{n\pi}{4}\big)\Big).$

This is a rather amazing and beautiful result.

Next page - Links forward - Series for e