Tính tổng sau: $\Large S =\dfrac{1}{2} C_{n}^{0}-\dfrac{1}{4} C_{n}^{1

Tính tổng sau:

$\Large S =\dfrac{1}{2} C_{n}^{0}-\dfrac{1}{4} C_{n}^{1}+\dfrac{1}{6} C_{n}^{2}-\cdots+\dfrac{(-1)^{n}}{2(n+1)} C_{n}^{n}$

• A. $\Large \dfrac{1}{2(n+1)}$
• B. $\Large 2^n$
• C. $\Large 2^{n-1}$
• D. $\Large \dfrac{1}{n+1}$

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

Vì $\Large \dfrac{(-1)^{k}}{(k+1)} C_{n}^{k}=\dfrac{(-1)^{k}}{(n+1)} C_{n+1}^{k+1}$ nên:

$\Large S =\dfrac{1}{2(n+1)} \sum_{k=0}^{n}(-1)^{k} C_{n+1}^{k+1}$$\Large =\dfrac{-1}{2(n+1)}\left[\sum_{k=0}^{n+1}(-1)^{k} C_{n+1}^{k}-C_{n+1}^{0}\right]=\dfrac{1}{2(n+1)}$