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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.

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