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2 tháng 8 2015

\(1+\frac{1}{2}.\left(1+2\right)+\)\(\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)

=\(\frac{2}{2}+\frac{3}{2}+\frac{6}{3}+...+\frac{136}{16}\)

=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

=\(\frac{2+3+4+5+6+...+17}{2}\)=\(\frac{152}{2}=76\)

9 tháng 11 2018

Ta có:

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

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

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

\(=\frac{\frac{23.20}{2}}{2}=\frac{23.10}{2}=115\)

18 tháng 7 2016

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

Cho \(n\in\)N*.CMR:\(\frac{1}{n}.\left(1+2+...+n\right)=\frac{n+1}{2}\)

Ta có:\(\frac{1}{n}.\left(1+2+......+n\right)=\frac{1}{n}.\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\)

Ta có:\(1+\frac{1}{2}\left(1+2\right)+......+\frac{1}{20}.\left(1+2+.....+20\right)\)

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

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

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

\(=\frac{230}{2}=165\)

25 tháng 5 2016

Mình mới học lớp 5

25 tháng 5 2016

mình ko trả lời được đâu nha!

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

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

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

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

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

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

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

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