14. Find
$$
2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}
$$

the value.
14. Solution 1: Since $k \mathrm{C}_{n}^{k}=n \mathrm{C}_{n-1}^{k-1}$, we have:
$$
\begin{array}{l}
\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-1}=n 2^{n-1} \\
\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1) \mathrm{C}_{n-1}^{k-1}+n \sum_{k=1}^{n} \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=2}^{n}(n-1) \mathrm{C}_{n-2}^{k-2}+n 2^{n-1} \\
=n(n-1) 2^{n-2}+n 2^{n-1}=n(n+1) 2^{n-2} \\
\sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k^{2} n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1)^{2} \mathrm{C}_{n-1}^{k-1}+n \sum_{k=1}^{n}(2 k-1) \mathrm{C}_{n-1}^{k-1} \\
=n(n-1) n 2^{n-3}+2 n(n+1) 2^{n-2}-n 2^{n-1} \\
=n^{2}(n+3) 2^{n-3}
\end{array}
$$

Thus, $2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}$
$$
\begin{array}{l}
=\left[2 n^{2}(n+3)-3 n \times 2 n(n+1)+n^{2} \times 4 n\right] 2^{n-3} \\
=0
\end{array}
$$

Solution 2: The original expression $=\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}\left(2 k^{2}-3 n k+n^{2}\right)$
$$
=\sum_{k=1}^{n} k(n-k)(n-2 k) \mathrm{C}_{n}^{k} \text {. }
$$

Let $M_{k}=k(n-k)(n-2 k) \mathrm{C}_{n}^{k}$, then $M_{n-k}=-M_{k}$.
When $n$ is even, the $\frac{n}{2}$-th term and the $n$-th term are 0; when $n$ is odd, the $n$-th term is 0.
In summary, the original expression equals 0.