K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

3 tháng 4 2018

Áp dụng công thức: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

=> \(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+...+20\right)\)

\(C=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{20}.\frac{20.21}{2}=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

=> \(C=\frac{1}{2}\left(2+3+4+...+21\right)=\frac{230}{2}=115\)

Đáp số: C=115

3 tháng 11 2016

C=\(\frac{483}{2}\)

20 tháng 2 2017

C=\(1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+....+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)

C=\(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{20}.\frac{20.21}{2}\)

C=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{21}{2}\)

C=\(\frac{2+3+.....+21}{2}\)

C=230

16 tháng 1 2019

\(\Rightarrow B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+....+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)

\(\Rightarrow B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{20}.\frac{21.20}{2}\)

\(\Rightarrow B=1+\frac{1}{2}.3+\frac{4}{2}+...+\frac{21}{2}\)

\(\Rightarrow B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

\(\Rightarrow B=\frac{2+3+4+...+21}{2}=...\)

Good Clever

16 tháng 1 2019

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{20}\cdot\frac{20\cdot21}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{1+2+3+....+21}{2}-\frac{1}{2}\)

\(=\frac{21\cdot22}{2}\cdot\frac{1}{2}-\frac{1}{2}\)

\(=\frac{1}{2}\left(\frac{21\cdot22}{2}-1\right)\)

\(=230\cdot\frac{1}{2}\)

4 tháng 7 2017

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)

\(=\frac{1}{n+1}\)

\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+5+...+21}{2}=115\)

16 tháng 10 2015

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

\(=\frac{1.2}{2}+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)