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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
a)
\(\begin{array}{l}{\left( {1 + \frac{1}{2} - \frac{1}{4}} \right)^2}.\left( {2 + \frac{3}{7}} \right)\\ = {\left( {\frac{4}{4} + \frac{2}{4} - \frac{1}{4}} \right)^2}.\left( {\frac{{14}}{7} + \frac{3}{7}} \right)\\ = {\left( {\frac{5}{4}} \right)^2}.\frac{{17}}{7}\\ = \frac{{25}}{{16}}.\frac{{17}}{7}\\ = \frac{{425}}{{112}}\end{array}\)
b)
\(\begin{array}{l}4:{\left( {\frac{1}{2} - \frac{1}{3}} \right)^3}\\ = 4:{\left( {\frac{3}{6} - \frac{2}{6}} \right)^3}\\ = 4:{\left( {\frac{1}{6}} \right)^3}\\ = 4:\frac{1}{{216}}\\ = 4.216\\ = 864\end{array}\)
Ta có: 1+2+3+...+n=\(\frac{n\left(n+1\right)}{2}\)
=> \(1=\frac{1x2}{2};\frac{1}{2}\left(1+2\right)=\frac{2x3}{2x2};\frac{1}{3}\left(1+2+3\right)=\frac{3x4}{2x3};\)\(;\frac{1}{4}\left(1+2+3+4\right)=\frac{4x5}{2x4};...;\frac{1}{20}\left(1+2+3+...+20\right)=\frac{20x21}{2x20}\)
=> \(B=\frac{1x2}{2}+\frac{2x3}{2x2}+\frac{3x4}{2x3}+\frac{4x5}{2x4}+...+\frac{20x21}{2x20}\)
=> \(B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
=> \(B=\frac{1}{2}\left(2+3+4+5+...+21\right)=\frac{1}{2}\left(\frac{21.22}{2}-1\right)\)
=> \(B=\frac{230}{2}=115\)
Đáp số: B=115
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{20}.\left(1+...+20\right).\)
\(=1+\frac{3}{2}+\frac{6}{3}+\frac{10}{4}+...+\frac{210}{20}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=\frac{230}{2}=115\)
\(\Rightarrow B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+....+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{20}.\frac{21.20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.3+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2+3+4+...+21}{2}=...\)
Good Clever
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{20}\cdot\frac{20\cdot21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{1+2+3+....+21}{2}-\frac{1}{2}\)
\(=\frac{21\cdot22}{2}\cdot\frac{1}{2}-\frac{1}{2}\)
\(=\frac{1}{2}\left(\frac{21\cdot22}{2}-1\right)\)
\(=230\cdot\frac{1}{2}\)
Bí
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=\frac{1.2}{2}+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)
Ta đã biết công thức: \(1+2+3+......+n-1+n=\frac{n\left(n+1\right)}{2}\).
Vậy:\(1+2=\frac{2\left(2+1\right)}{2}=\frac{2.3}{2}\); \(1+2+3=\frac{3\left(3+1\right)}{2}=\frac{3.4}{2}.\)a có:
Thay vào bài toán ta có:
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{20}\left(1+2+3+....+20\right)\)
\(=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+....+\frac{1}{20}.\frac{20.21}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{21}{2}\)
\(=\frac{2+3+4+......+20+21}{2}=\frac{21\left(21+1\right)-1}{2}=\frac{461}{2}.\)
461/2