: Tính các tổng sau: ${{S}_{1}}=C_{n}^{0}+\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}+…+\dfrac{1}{n+1}C_{n}^{n}$ C. $\dfrac{{{2}^{n+1}}+1}{n+1}$

: Tính các tổng sau: ${{S}_{1}}=C_{n}^{0}+\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}+…+\dfrac{1}{n+1}C_{n}^{n}$
C. $\dfrac{{{2}^{n+1}}+1}{n+1}$
B. $\dfrac{{{2}^{n+1}}-1}{n+1}$
C. $\dfrac{{{2}^{n+1}}-1}{n+1}+1$
D. $\dfrac{{{2}^{n+1}}-1}{n+1}-1$

Chọn B
Ta có:
$\dfrac{1}{k+1}C_{n}^{k}=\dfrac{1}{k+1}\dfrac{n!}{k!(n-k)!}=\dfrac{1}{n+1}\dfrac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ (}n+1)-(k+1))!}$
$=\dfrac{1}{n+1}C_{n+1}^{k+1}$ (*)
$\Rightarrow {{S}_{1}}=\dfrac{1}{n+1}\sum\limits_{k=0}^{n}{C_{n+1}^{k+1}}=\dfrac{1}{n+1}\left( \sum\limits_{k=0}^{n+1}{C_{n+1}^{k}}-C_{n+1}^{0} \right)=\dfrac{{{2}^{n+1}}-1}{n+1}$.