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\(S=0.2+2.4+4.6+...+2n\left(2n+2\right)\)
\(6S=2.4.6+4.6.\left(8-2\right)+...+2n\left(2n+2\right)\left[\left(2n+4\right)-\left(2n-2\right)\right]\)
\(=2.4.6+4.6.8-2.4.6+...+2n\left(2n+2\right)\left(2n+4\right)-\left(2n-2\right).2n.\left(2n+2\right)\)
\(=2n\left(2n+2\right)\left(2n+4\right)\)
Suy ra \(S=\frac{2n\left(2n+2\right)\left(2n+4\right)}{6}\)
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\(S=\frac{1}{2022}-\left(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\right)\)
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\)
\(A=\frac{5}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2022-2020}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)\)
\(S=\frac{1}{2022}-A=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=-\frac{1262}{1011}\)
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\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+....+\frac{1}{98\cdot100}\)
\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+.......+\frac{2}{98\cdot100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)
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Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)