1/50*51+1/51*52+...........+1/99*100
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Sửa đề: \(S=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)
Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)
Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)
Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)
Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)
Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)
=>\(S<\frac56\)
Ta có: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\)
\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{99}+\frac{1}{100}-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\)
\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{99}+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)
=>\(A-B=-\frac{1}{50}\)
\(\frac{1}{50\times51}+\frac{1}{51\times52}+...+\frac{1}{99\times100}\)
\(=\frac{51-50}{50\times51}+\frac{52-51}{51\times52}+...+\frac{100-99}{99\times100}\)
\(=\frac{1}{50}-\frac{1}{51}+\frac{1}{51}-\frac{1}{52}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{50}-\frac{1}{100}=\frac{1}{100}\)