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\(\frac{1}{1x3x5}+\frac{1}{5x7x9}+\frac{1}{9x11x13}+.....+\frac{1}{49x51x53}=\)
\(1-\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{49}-\frac{1}{51}-\frac{1}{53}=\)
\(1-\frac{1}{3}-\frac{1}{7}-....-\frac{1}{51}-\frac{1}{53}=\)
\(A=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{10}-\frac{1}{100}\)
\(A=\frac{9}{100}\)
\(A=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
\(=\frac{9}{100}\)
Ta có: \(\frac{1}{4\times6}=\frac{1}{4\times1\times3\times2}=\frac{1}{4\times3\times1\times2}\)
\(\frac{1}{8\times9}=\frac{1}{4\times2\times3\times3}=\frac{1}{4\times3\times2\times3}\)
\(\frac{1}{12\times12}=\frac{1}{4\times3\times3\times4}\)
\(\frac{1}{16\times15}=\frac{1}{4\times4\times3\times5}=\frac{1}{4\times3\times4\times5}\)......
\(\frac{1}{2680\times2013}=\frac{1}{4\times670\times3\times671}\)
Do đó:
\(M=\frac{1}{4\times3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{670\times671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{670}-\frac{1}{671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{671}\right)=\frac{1}{12}\times\frac{670}{671}=\frac{335}{4026}\)
Vậy \(M=\frac{335}{4026}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
\(M=\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)-....-\left(1-\frac{1}{200}\right)\)
\(M=-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-.....-\frac{1}{200}\right)=-\frac{1}{2}\left(1-\frac{1}{2}+...-\frac{1}{100}\right)\)
Xét:
\(S=1-\frac{1}{2}+....-\frac{1}{100}.S=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+....+\frac{1}{100}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\Rightarrow M=-\frac{1}{2}\left(\frac{1}{51}+....+\frac{1}{100}\right)\)
N:M=-2