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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!)
\(B=\left(1+2+3+....+10\right)+\left(1+2+3+...+10\right)+\left(\frac{1}{2}.10\right)\)
\(B=55+55+5=115\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{20}.\left(1+2+...+20\right)\)
\(=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(3+1\right).3}{2}+...+\frac{1}{20}.\frac{\left(20+1\right).20}{2}\)
\(=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{20+1}{2}\)
\(=1+\frac{1}{2}.\left(1+2+1+3+...+20+1\right)\)
\(=1+\frac{1}{2}.\left[\left(1+1+...+1\right)+\left(1+2+3+...+20\right)\right]\)
\(=1+\frac{1}{2}.\left[20+\frac{\left(20+2\right).19}{2}\right]\)
\(=1+\frac{1}{2}.\left[20+\frac{22.19}{2}\right]\)
\(=1+\frac{1}{2}.\left[20+11.19\right]\)
\(=1+\frac{1}{2}.\left[20+209\right]\)
\(=1+\frac{1}{2}.229\)
\(=\frac{2}{2}+\frac{229}{2}\)
\(=\frac{231}{2}\)
Tham khảo nhé~
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+....+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+.....+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+....+\frac{21}{2}\)
\(=\frac{2+3+.....+21}{2}=\frac{230}{2}=115\)
\(1+\frac{1}{2}.\left(1+2\right)+\)\(\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
=\(\frac{2}{2}+\frac{3}{2}+\frac{6}{3}+...+\frac{136}{16}\)
=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
=\(\frac{2+3+4+5+6+...+17}{2}\)=\(\frac{152}{2}=76\)
Ta có:
\(1+\frac{1}{2}\left(1+2\right)+..........+\frac{1}{20}\left(1+2+3+.......+20\right)\)
\(=1+\frac{1}{2}\left(\frac{3.2}{2}\right)+\frac{1}{3}\left(\frac{4.3}{2}\right)+........+\frac{1}{20}\left(\frac{21.20}{2}\right)\)
\(=1+\frac{3}{2}+\frac{4}{2}+..........+\frac{21}{2}=\frac{2+3+4+........+21}{2}\)
\(=\frac{\frac{23.20}{2}}{2}=\frac{23.10}{2}=115\)
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
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)