<s>[INST] Assuming that $|z|<1$, calculate: $\sum_{n=2}^{\infty}(n^2-3n+2){z^{n-1}}$ The answer is: $\frac{-2z^2}{(1-z)^{3}}$
When I do it, I get the answer without the minus sign.
So far, I got:
$$
\begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^n
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^n \\
&= z^3 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\
&= z^3 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}
\end{split}
$$
 [/INST] \begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^{n-1}
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\
&= z^2 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-3} \\
&= z^2 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}\\
&= z^2 \frac{d^2}{dz^2} \sum_{n=0}^{\infty}z^{n}\\
&= z^2 \frac{d^2}{dz^2} \frac{1}{1-z}\\
&= z^2 \frac{2}{(1-z)^3}
\end{split}
This proof holds for all complex $z$ with $|z|<1$ which means that, in particular, this also holds for all real $z$ with $0<z<1$.  From the task setting itself, in this case the answer must be positive, since only positive terms are added. So if the result is claimed to be $ \frac{-2 z^2}{(1-z)^3}$, this is false.
</s>