［答案］$\frac{1}{n+1}\left[\left(\frac{3}{2}\right)^{n+1}-1\right]$
［解析］法一：注意到信息中的积分算法，所以逆写可得
$\mathrm{C}_{\mathrm{n}}^{0} \times \frac{1}{2}+\frac{1}{2} \mathrm{C}_{\mathrm{n}}^{1} \times\left(\frac{1}{2}\right)^{2}+\frac{1}{3} \mathrm{C}_{\mathrm{n}}^{2} \times\left(\frac{1}{2}\right)^{3}+\cdots+\frac{1}{n+1} \mathrm{C}_{\mathrm{n}}^{\mathrm{n}} \times\left(\frac{1}{2}\right)^{n+1}$
$=\int_{0}^{\frac{1}{2}} \mathrm{C}_{\mathrm{n}}^{0} x^{0} d x+\int_{0}^{\frac{1}{2}} \mathrm{C}_{\mathrm{n}}^{1} x^{1} d x+\int_{0}^{\frac{1}{2}} \mathrm{C}_{\mathrm{n}}^{2} x^{2} d x+\cdots+\int_{0}^{\frac{1}{2}} \mathrm{C}_{\mathrm{n}}^{\mathrm{n}} x^{n} d x\left({ }^{*}\right)$
$\because \mathrm{C}_{\mathrm{n}}^{0} x^{0}+\mathrm{C}_{\mathrm{n}}^{1} x^{1}+\mathrm{C}_{\mathrm{n}}^{2} x^{2}+\mathrm{C}_{\mathrm{n}}^{3} x^{3}+\cdots+\mathrm{C}_{\mathrm{n}}^{\mathrm{n}} x^{n}=(1+x)^{n}$
$$
\therefore(*)=\int_{0}^{\frac{1}{2}}(1+x)^{n} d x=\left.\frac{1}{n+1}(1+x)^{n+1}\right|_{0} ^{\frac{1}{2}}=\frac{1}{n+1}\left[\left(\frac{3}{2}\right)^{n+1}-1\right]
$$
法二：考虑组合恒等式 $\frac{1}{k+1} C_{n}^{k}=\frac{1}{n+1} C_{n+1}^{k+1}$ 故直接可得
$$
\begin{aligned}
C_{n}^{0} & \times \frac{1}{2}+\frac{1}{2} C_{n}^{1} \times\left(\frac{1}{2}\right)^{2}+\frac{1}{3} C_{n}^{2} \times\left(\frac{1}{2}\right)^{3}+\cdots+\frac{1}{n+1} C_{n}^{n} \times\left(\frac{1}{2}\right)^{n+1} \\
& =\frac{1}{n+1}\left[C_{n+1}^{1} \frac{1}{2}+C_{n+1}^{2} \times\left(\frac{1}{2}\right)^{2}+C_{n+1}^{3} \times\left(\frac{1}{2}\right)^{3}+\cdots+C_{n+1}^{n+1} \times\left(\frac{1}{2}\right)^{n+1}\right] \\
& =\frac{1}{n+1}\left[\left(1+\frac{1}{2}\right)^{n+1}-1\right]=\frac{1}{n+1}\left[\left(\frac{3}{2}\right)^{n+1}-1\right]
\end{aligned}
$$
［ 考点定位］此题的立意在类比应用，巧㠺的逆向构造考查了学生应用信息的能力。难度比较大。不过如果参加竞赛或者熟悉 $\frac{1}{k+1} C_{n}^{k}=\frac{1}{n+1} C_{n+1}^{k+1}$ 恒等式也就比较容易了。
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