Question

19．计算：${C}_{n}^{0}×\frac{1}{2}$+$\frac{1}{2}$${C}_{n}^{1}$×（$\frac{1}{2}$）²+$\frac{1}{3}$${C}_{n}^{2}$×（$\frac{1}{2}$）³+…+$\frac{1}{n+1}$${C}_{n}^{n}$×（$\frac{1}{2}$）ⁿ⁺¹=$\frac{1}{n+1}[（\frac{3}{2}）^{n+1}-1]$．．

Answer 1

分析 因为${C}_{n}^{0}+{C}_{n}^{1}x+{C}_{n}^{2}{x}^{2}+…{C}_{n}^{n}{x}^{n}=（1+x）^{n}$，两边进行0到$\frac{1}{2}$的定积分，然后求定积分值即可．

解答 解：因为${C}_{n}^{0}+{C}_{n}^{1}x+{C}_{n}^{2}{x}^{2}+…{C}_{n}^{n}{x}^{n}=（1+x）^{n}$，
两边对x定积分，即${∫}_{0}^{\frac{1}{2}}{C}_{n}^{0}dx+{∫}_{0}^{\frac{1}{2}}{C}_{n}^{1}xdx$+…+${∫}_{0}^{\frac{1}{2}}{C}_{n}^{n}{x}^{n}dx$=${∫}_{0}^{\frac{1}{2}}{（1+x）}^{n}dx$，
所以${C}_{n}^{0}×\frac{1}{2}$+$\frac{1}{2}$${C}_{n}^{1}$×（$\frac{1}{2}$）²+$\frac{1}{3}$${C}_{n}^{2}$×（$\frac{1}{2}$）³+…+$\frac{1}{n+1}$${C}_{n}^{n}$×（$\frac{1}{2}$）ⁿ⁺¹=$\frac{1}{n+1}（1+x）^{n+1}{|}_{0}^{\frac{1}{2}}$=$\frac{1}{n+1}[（\frac{3}{2}）^{n+1}-1]$．
故答案为：$\frac{1}{n+1}[（\frac{3}{2}）^{n+1}-1]$．

点评 本题考查了二项式定理的灵活运用；关键是发现已知式子是${C}_{n}^{0}+{C}_{n}^{1}x+{C}_{n}^{2}{x}^{2}+…{C}_{n}^{n}{x}^{n}=（1+x）^{n}$在0到$\frac{1}{2}$的定积分．