Find the generating function for $ a_n = n^4$ .

Question

Find the generating function for $a_n = n^4$.

How should I approach this question? I have tried to go $1,16,81\dots$ and generate the function with the series but no idea how to do so.

Answer 1

Notice

$$ \sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$

Take a derivative to get

$$\sum_{n=0}^\infty (n+1) x^{n} = \frac{1}{(1-x)^2}.$$

Now we get

$$ \frac{1}{(1-x)^2} - \frac{1}{1-x} = \sum_{n=0}^\infty n x^n + \sum_{n=0}^\infty x^n - \sum_{n=0}^\infty x^n = \sum_{n=0}^\infty n x^n.$$

Keep going like this.

Answer 2

\begin{align} \sum_{n \ge 0} n^4 z^n &=\sum_{n \ge 0} \left(\binom{n}{1}+14\binom{n}{2}+36\binom{n}{3}+24\binom{n}{4}\right) z^n \\ &=\sum_{n \ge 1} \binom{n}{1} z^n + 14\sum_{n \ge 2}\binom{n}{2} z^n + 36\sum_{n \ge 3}\binom{n}{3} z^n + 24\sum_{n \ge 4}\binom{n}{4} z^n \\ &=z \sum_{n \ge 0} \binom{n+1}{1} z^n + 14z^2\sum_{n \ge 0}\binom{n+2}{2} z^n + 36z^3\sum_{n \ge 0}\binom{n+3}{3} z^n + 24z^4\sum_{n \ge 0}\binom{n+4}{4} z^n \\ &=\frac{z}{(1-z)^2} + \frac{14z^2}{(1-z)^3} + \frac{36z^3}{(1-z)^4} + \frac{24z^4}{(1-z)^5} \\ &= \frac{z + 11 z^2 + 11 z^3 + z^4}{(1-z)^5} \end{align}

Answer 3

The simplest way to get the answer is to insert the terms 1,16,81,256 into the OEIS query form and read the generating function there: https://oeis.org/A000583 .