Let $$f(x) = \sum\_{n\geq 0} a\_n x^n = \frac{P(x)}{(1-x)^d}$$ be a rational function.

(a) Prove: There is a polynomial $P\_2(x)$ so $$\sum\limits\_{n\geq 0} a\_{2\_n} x^n = \frac{P\_2(x)}{(1-x)^d}$$

(b) Let $r \geq 1 \in \mathbb{N}$. Show that an polynomial $P\_r$ exists so that $$\sum\_{n\geq 0} a\_{rn} x^n = \frac{P\_r(x)}{(1-x)^d}$$

Hint: Use the $r$th roots of unity which are defined by $\exp\left(\large \frac{2\pi ik}{r}\right), 0 \leq k \leq r - 1$.

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(a) I don't know what this *d* is about (and no one else did). Might be an absolute term. 

As $f(x)$ is a rational function, it can be defined as a fraction of two polynomials $\large \frac{P(x)}{Q(x)}$. But that is unfortunately all I know about this. 

Could you please help me going on?

(b) I don't know how the $r$th roots of unity (and therefore numbers $x$ for which applies: $x^r = 1$) can help me solving this? I don't find any approach.

Could you please help me a bit?

Thanks in advance!