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

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

Chọn A
Ta có: $S=\dfrac{1}{2}\left( C_{n}^{0}-\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}-…+\dfrac{{{(-1)}^{n}}}{n+1}C_{n}^{n} \right)$
Vì $\dfrac{{{(-1)}^{k}}}{k+1}C_{n}^{k}=\dfrac{{{(-1)}^{k}}}{n+1}C_{n+1}^{k+1}$ nên:$S=\dfrac{1}{2(n+1)}\sum\limits_{k=0}^{n}{{{(-1)}^{k}}C_{n+1}^{k+1}}$
$=\dfrac{-1}{2(n+1)}\left( \sum\limits_{k=0}^{n+1}{{{(-1)}^{k}}C_{n+1}^{k}}-C_{n+1}^{0} \right)=\dfrac{1}{2(n+1)}$.