: Tính tổng sau: $S=\frac{1}{2}C_{n}^{0}-\frac{1}{4}C_{n}^{1}+\frac{1}{6}C_{n}^{3}-\frac{1}{8}C_{n}^{4}+…+\frac{{{(-1)}^{n}}}{2(n+1)}C_{n}^{n}$

: Tính tổng sau: $S=\frac{1}{2}C_{n}^{0}-\frac{1}{4}C_{n}^{1}+\frac{1}{6}C_{n}^{3}-\frac{1}{8}C_{n}^{4}+…+\frac{{{(-1)}^{n}}}{2(n+1)}C_{n}^{n}$

C. $\frac{1}{2(n+1)}$

B. 1

C. 2

D. $\frac{1}{(n+1)}$

Hướng dẫn

Chọn A

Ta có: $S=\frac{1}{2}\left( C_{n}^{0}-\frac{1}{2}C_{n}^{1}+\frac{1}{3}C_{n}^{2}-…+\frac{{{(-1)}^{n}}}{n+1}C_{n}^{n} \right)$

Vì $\frac{{{(-1)}^{k}}}{k+1}C_{n}^{k}=\frac{{{(-1)}^{k}}}{n+1}C_{n+1}^{k+1}$ nên:$S=\frac{1}{2(n+1)}\sum\limits_{k=0}^{n}{{{(-1)}^{k}}C_{n+1}^{k+1}}$

$=\frac{-1}{2(n+1)}\left( \sum\limits_{k=0}^{n+1}{{{(-1)}^{k}}C_{n+1}^{k}}-C_{n+1}^{0} \right)=\frac{1}{2(n+1)}$.