Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.
I would love some help with this.

So your sum looks like
$$
s(x) = 2 + \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k.
$$
One approach is to expand the sum to
$$
\begin{split}
\sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k
 &= \sum_{k=1}^\infty \frac{k^2+2k+1}{k} x^k \\
 &= \sum_{k=1}^\infty kx^k
    + 2 \sum_{k=1}^\infty x^k
    + \sum_{k=1}^\infty \frac{x^k}{k}.
\end{split}
$$
The first sum you can get by differentiating a geometric series:
$$
\begin{split}
g(x)    &= \sum_{k=0}^\infty x^k\\
g'(x)   &= \sum_{k=1}^\infty k x^{k-1}\\
x g'(x) &= \sum_{k=1}^\infty k x^k
\end{split}
$$
The second sum is a geometric series. Differentiate the third sum to get a geometric series:
$$
\begin{split}
f(x)  &= \sum_{k=1}^\infty \frac{x^k}{k}\\
f'(x) &= \sum_{k=1}^\infty x^{k-1}
\end{split}
$$
which you can now find...
Just make sure to be careful about the term where $k=0$ everywhere...
