17. If $n$ is a positive integer, find the formula for the sum $\binom{n+1}{1}$
$$
+\binom{n+2}{2} 2^{2}+\binom{n+3}{3} 3^{2}+\cdots+\binom{2 n}{n} n^{2}
$$
17. Consider a chess club with $2 n+1$ players, ranked from 1 to $2 n$ +1. Select $n+1$ players to form a team for an upcoming competition with a competitive club. Additionally, choose one captain and one coach from the remaining $n$ players. The same player can hold both positions, but each captain's rank must be at least one higher than any team member.

If $k$ is the lowest-ranked team member, then $n+2 \leqslant k \leqslant 2 n$ +1. For each such $k$, the remaining $n$ team members can be chosen in $\binom{k-1}{n}=\binom{k-1}{k-1-n}$ ways, and the captain can be chosen in $(k-1-n)^{2}$ ways. Therefore, the total number of selection methods is $\binom{n+1}{1}+2^{n}\binom{n+2}{2}+\cdots$ $n^{2}\binom{2 n}{n}$, which is the required sum.

There are $\binom{2 n+1}{n+1}$ ways to choose the team members, and $n^{2}$ ways to choose the captain, but we must exclude selections where at least one captain's rank is higher than all team members. Assuming the two captain positions are held by the same player, there are $\binom{2 n+1}{n+2}$ ways to choose the team, plus this captain, and the lowest-ranked player among these $n+2$ players will be the sole captain. If the captain and coach are different players, there are $\binom{2 n+1}{n+3}$ ways to choose the team plus these two captains. If the lowest-ranked player among these $n+3$ players only serves as one captain, the other positions for the remaining $n+2$ players must be determined. Therefore, the total number of methods is
$$
\begin{array}{c}
n^{2}\binom{2 n+1}{n+1}-\binom{2 n+1}{n+2}-2(n+1) \\
\cdot\binom{2 n+1}{n+3}=\frac{n(n+1)^{3}\binom{2 n+1}{n+1}}{(n+2)(n+3)},
\end{array}
$$
which is the formula for the required sum.