Calculate $S_{n}=\sum_{k=1}^{n} k^{2}\binom{n}{k}$
We start by calculating

$$
A_{n}=\sum_{k=1}^{n} k\binom{n}{k}
$$

Among $n$ balls, there is one red ball and any number of blue balls. First method:

1. Choose $k$ colored balls: $\binom{n}{k}$ possible choices.
2. Choose the red ball among these $k$ balls: $k$ possible choices.

In total, by considering all possible $k$, we obtain $A_{n}$. Second method:

1. Choose the red ball: $n$ possible choices.
2. Choose blue balls from the remaining $n-1$ balls: $2^{n-1}$ possible choices.

Thus, $A_{n}=n 2^{n-1}$. Conclusion

$$
\sum_{k=1}^{n} k\binom{n}{k}=n 2^{n-1}
$$

Now we calculate $B_{n}=\sum_{k=1}^{n} k(k-1)\binom{n}{k}$. Among $n$ balls, there is one red ball, one green ball, and any number of blue balls. First method:

1. Choose $k$ colored balls: $\binom{n}{k}$ possible choices.
2. Choose the red ball among these $k$ balls: $k$ possible choices.
3. Choose the green ball among the remaining $k-1$ balls: $k-1$ possible choices. And we obtain $B_{n}$ by considering all possible $k$. Second method:
4. Choose the red ball: $n$ possible choices.
5. Choose the green ball: $n-1$ possible choices.
6. Choose blue balls from the remaining $n-2$ balls: $2^{n-2}$ possible choices. Thus, $B_{n}=n(n-1) 2^{n-2}$. Conclusion

$$
\begin{aligned}
S_{n} & =\sum_{k=1}^{n}(k(k-1)+k)\binom{n}{k} \\
& =B_{n}+A_{n} \\
& =n 2^{n-1}+n(n-1) 2^{n-2} \\
& =n(n+1) 2^{n-2}
\end{aligned}
$$