<s>[INST] Geometric series and polynomials 
i) Find a generating function expression of a sequence with terms 
  $$d_n=\sum_{p=0}^n p^3$$
  using operations on the geometric series $\sum_{n\geq 0} x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$. 

for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
 [/INST] Yes, the generating function is correct: just apply the operator $x\frac{d}{dx}$ to $\sum_{n\geq 0} x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $\sum_{k=0}^n$):
$$f(x)=\frac{x(1+4x+x^2)}{(1-x)^5}$$
Now, in order to find a polynomial formula for $\sum_{k=0}^n k^3$ we have to extract the coefficient of $x^n$,
$$\begin{align}\sum_{k=0}^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)\cdot(1-x)^{-5}\\&=
[x^{n-1}](1-x)^{-5}+4[x^{n-2}](1-x)^{-5}+
[x^{n-3}](1-x)^{-5}\\
&=(-1)^{n-1}\binom{-5}{n-1}+4(-1)^{n-2}\binom{-5}{n-2}+(-1)^{n-3}\binom{-5}{n-3}\\
&=\binom{n+3}{4}+4\binom{n+2}{4}+\binom{n+1}{4}
=\frac{n^2(n+1)^2}{4}
\end{align}$$
where for the expansion of $(1-x)^{-5}$ we used the Newton's generalized binomial theorem and 
$$\binom{-r}k =(-1)^{k}\binom{r+k-1}{r-1}.$$
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