Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fractions $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}$.
I know the series representations of these functions $(1+x)^{n}$ and $\frac{1}{1-x}$, and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it.

Note $$\frac{1}{1 + x + \cdots + x^{n-1}} = \frac{1-x}{1-x^n}.$$  Since we know $$\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{k=0}^\infty z^k, \quad |z| < 1,$$ we have $$\frac{1-x}{1-x^n} = (1-x)\sum_{k=0}^\infty (x^n)^k = (1-x)(1+x^n + x^{2n} + x^{3n} + \cdots).$$  We conclude that the desired series expansion is $$1 - x + x^n - x^{n+1} + x^{2n} - x^{2n+1} + x^{3n} - x^{3n+1} + \cdots,$$ for integers $n > 1$.  Your case is $n = 4$.

It is also worth pointing out that because of the restriction $|z| < 1$ for the geometric series expansion we used above, the given series expansion does not converge even when the original function is well-defined.  This is addressed by performing the series expansion about $z = \infty$ rather than $z = 0$.  We do this by writing $$\frac{1}{1-z} = \frac{z^{-1}}{z^{-1} - 1} = -z^{-1} \sum_{k=0}^\infty z^{-k} = -\sum_{k=1}^\infty z^{-k}, \quad |z| > 1,$$ hence $$\frac{1-x}{1-x^n} = -(1-x) \sum_{k=1}^\infty (x^n)^{-k} = x^{-n+1} - x^{-n} + x^{-2n+1} - x^{-2n} + x^{-3n+1} - x^{-3n} + \cdots, \quad |x| > 1.$$
