Phạm Ngọc Huy
New member
: 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. $\dfrac{1}{(n+1)}$
C. $\dfrac{1}{2(n+1)}$
B. 1
C. 2
D. $\dfrac{1}{(n+1)}$