Tháng Ba 29, 2024

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

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

C. $\frac{{{2}^{n+1}}+1}{n+1}$

B. $\frac{{{2}^{n+1}}-1}{n+1}$

C. $\frac{{{2}^{n+1}}-1}{n+1}+1$

D. $\frac{{{2}^{n+1}}-1}{n+1}-1$

Hướng dẫn

Chọn B

Ta có:

$\frac{1}{k+1}C_{n}^{k}=\frac{1}{k+1}\frac{n!}{k!(n-k)!}=\frac{1}{n+1}\frac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ (}n+1)-(k+1))!}$

$=\frac{1}{n+1}C_{n+1}^{k+1}$ (*)

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