Question

19．計(jì)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?yàn)?{C}_{n}^{0}+{C}_{n}^{1}x+{C}_{n}^{2}{x}^{2}+…{C}_{n}^{n}{x}^{n}=（1+x）^{n}$，兩邊進(jìn)行0到$\frac{1}{2}$的定積分，然后求定積分值即可．

解答 解：因?yàn)?{C}_{n}^{0}+{C}_{n}^{1}x+{C}_{n}^{2}{x}^{2}+…{C}_{n}^{n}{x}^{n}=（1+x）^{n}$，
兩邊對(duì)x定積分，即${∫}_{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]$．
故答案為：$\frac{1}{n+1}[（\frac{3}{2}）^{n+1}-1]$．

點(diǎn)評(píng) 本題考查了二項(xiàng)式定理的靈活運(yùn)用；關(guān)鍵是發(fā)現(xiàn)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$\frac{1}{2}$的定積分．