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?
 A: 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 
$$\binom{-r}k =(-1)^{k}\binom{r+k-1}{r-1}.$$
A: Consider $$S_n=\sum_{k=0}^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3\sum_{k=0}^n k(k-1)(k-2)x^{k-3}+3x^2\sum_{k=0}^n k(k-1)x^{k-2}+x\sum_{k=0}^n kx^{k-1}$$ that is to say
$$S_n=x^3\left(\sum_{k=0}^n x^{k} \right)'''+3x^2\left(\sum_{k=0}^n x^{k} \right)''+x\left(\sum_{k=0}^n x^{k} \right)'$$
When finished, consider the limit when $x\to 1$ to get the beautiful result.
A: Using 
$$\sum_{k=0}^{n} x^k = \frac{1 - x^{n+1}}{1-x}$$
then by applying the operator $\delta = x D$, $D = \frac{d}{dx}$, leads to
$$\sum_{k=0}^{n} k^{3} \, x^{k} = \frac{x}{(1-x)^4} \, ( 1 + 4x + x^2 - (n+1)^3 \, x^n + (3n^3 + 6n^2 - 4) \, x^{n+1} - (3n^3 + 3n^2 - 3n +1) \, x^{n+2} + n^3 \, x^{n+3}).$$
By taking the limit as $x \to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$\sum_{k=0}^{n} k^3 = \frac{n^2 (n+1)^2}{4}.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$\sum_{n=0}^{\infty} \frac{n^2 (n+1)^2}{4} \, x^{n} = \frac{x(1+4 x + x^2)}{(1-x)^5}$$
which matches a previously stated polynomial by the proposer.
