Example 14 Find $\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1}$.
Solution: From $(1+x)^{11}=\sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k}$, we have
$$
\begin{array}{l}
\int_{0}^{1}(1+x)^{11} \mathrm{~d} x=\int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x . \\
\text { and } \int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x \\
=\left.\left(\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k} x^{k+1}}{k+1}\right)\right|_{0} ^{1}=\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1} \text {, } \\
\int_{0}^{1}(1+x)^{11} \mathrm{~d} x=\left.\frac{(1+x)^{12}}{12}\right|_{0} ^{1}=\frac{1365}{4} \text {. } \\
\end{array}
$$

Therefore, $\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1}=\frac{1365}{4}$.
Similarly, if we first differentiate the expansion of $(1+x)^{n}$, then multiply both sides by $x$, and differentiate again, and finally set $x=1$, we can prove Example 1; if we differentiate $(1+x)^{n}$ twice and then set $x=1$, we can prove the following conclusion
$$
\begin{array}{l}
2 \times 1 \times \mathrm{C}_{n}^{2}+3 \times 2 \times \mathrm{C}_{n}^{3}+\cdots+n(n-1) \mathrm{C}_{n}^{n} \\
=n(n-1) 2^{n-2} .
\end{array}
$$