Trying to find the closed form for the nth term of $\frac{1}{1-x^4}$

Question

I know that $\frac{1}{1-x^4}$ is the generating function for the sequence (1, 0, 0, 0, 1, 0, 0, 0, 1, ...) I don't know how to find the closed form for the nth term though. Itried messing around with $(1 + (-1)^n) / 2$ but couldn't come up with anything. My book doesn't use mod anywhere in the chapter either so I'm assuming it can be solved without it.

Answer 1

$$(1+(-1)^n) \cdot \left(1 + (-1)^{n (1 + (-1)^n)/4}\right)/4$$

But really, is this actually what's requested? This form isn't very useful.

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If you don't mind complex numbers, to expand on on coffeemath's answer below, in general $\sum_{j=0}^k e^{2\pi i jn/k}$ is $k$ if $k|n$ and $0$ otherwise. In the case $k = 4$ this specializes to $1^n + i^n + (-1)^n + (-i)^n$ equals $4$ if $4|n$ and $0$ otherwise.

Answer 2

Another formula using the powers of $i=\sqrt{-1}$ is $$a_n=\frac{1+i^n+(-1)^n+(-i)^n}{4}.$$ This has to do with the sum of $4$th roots of unity being zero, and the fact that when $n$ is a multiple of $4$ the terms on top are all $1.$