\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
ta có: \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
...
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
Vậy \(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
vậy A = 49/100
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