\(t=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(t=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(t=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(t=\frac{99}{100}\)
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