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

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

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

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

1 tháng 1 2020

115 nhé bạn

14 tháng 5 2017

Bài này hơi khó hiểu xíu. Thông cảm nha babe:v

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

\(B=1+\left(\frac{1}{2}+1\right)+2+\left(\frac{1}{2}+2\right)+3+\left(\frac{1}{2}+3\right)+.....+10+\left(\frac{1}{2}+10\right)\)(chỗ này là nhân phân phối vô đấy!)

\(B=\left(1+2+3+....+10\right)+\left(1+2+3+...+10\right)+\left(\frac{1}{2}.10\right)\)

\(B=55+55+5=115\)

11 tháng 8 2018

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

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

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

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

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

\(=1+\frac{1}{2}.\left[20+\frac{\left(20+2\right).19}{2}\right]\)

\(=1+\frac{1}{2}.\left[20+\frac{22.19}{2}\right]\)

\(=1+\frac{1}{2}.\left[20+11.19\right]\)

\(=1+\frac{1}{2}.\left[20+209\right]\)

\(=1+\frac{1}{2}.229\)

\(=\frac{2}{2}+\frac{229}{2}\)

\(=\frac{231}{2}\)

Tham khảo nhé~

11 tháng 8 2018

mk 

nha 

bn

ơi

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

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)