How to create a generating function / closed form from this recurrence? Let $f_n$ = $f_{n-1} + n + 6$ where $f_0 = 0$.
I know $f_n = \frac{n^2+13n}{2}$ but I want to pretend I don't know this. How do I correctly turn this into a generating function / derive the closed form?

In two answers, it is derived that the generating function is
$$
\begin{align}
\frac{7x-6x^2}{(1-x)^3}
&=x\left(\frac1{(1-x)^3}+\frac6{(1-x)^2}\right)\\
&=\sum_{k=0}^\infty(-1)^k\binom{-3}{k}x^{k+1}+6\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{k+1}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\left(\binom{-3}{k-1}+6\binom{-2}{k-1}\right)x^k\\
&=\sum_{k=1}^\infty\left(\binom{k+1}{k-1}+6\binom{k}{k-1}\right)x^k\\
&=\sum_{k=1}^\infty\left(\binom{k+1}{2}+6\binom{k}{1}\right)x^k\\
&=\sum_{k=1}^\infty\frac{k^2+13k}{2}x^k\\
\end{align}
$$
Therefore, we get the general term is $f_k=\dfrac{k^2+13k}{2}$
