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\(3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+100}\)
\(=3+3.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=3+3.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=3+3.\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=3+\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=3+\frac{3}{2}.\frac{99}{202}\)
\(=3+\frac{297}{404}\)
\(=\frac{1509}{404}\)
chỗ 3+3/2(1/6+..)
bn nhìn nhầm rồi
đáng lẽ: 3+(1/6+,.....) chứ nâk
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=\frac{1}{2}\left(2+3+...+101\right)=\frac{1}{2}\cdot\frac{100.103}{2}=25.103=2575\)
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+....+\frac{100}{2^{100}}\)
\(2A=2+\frac{3}{2^2}+\frac{4}{2^3}+....+\frac{100}{2^{99}}\)
\(2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
\(A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)+\frac{100}{2^{100}}\)\(\)
\(=1+\frac{3}{2^2}+\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\right)+\frac{100}{2^{100}}\)
\(=1+\frac{3}{2^2}+\frac{1}{2^2}-\frac{1}{2^{99}}+\frac{100}{2^{100}}\)
\(=1+\frac{4}{2^2}-\frac{2}{2^{100}}+\frac{100}{2^{100}}\)
\(=2-\frac{98}{2^{100}}=\frac{2^{101}-98}{2^{100}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\)
\(A-\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{1}{2}A=\left(1-\frac{1}{2^{101}}\right)\div\frac{1}{2}-\frac{100}{2^{101}}\)
\(=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(\Rightarrow A=\frac{\left(2^{101}-1\right)}{2^{99}}-\frac{100}{2^{100}}\)