Question

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+......+\frac{1}{97\cdot99}\)

Answer 1

đặt \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\) là A

A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

2A = 2 . (\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\))

2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

2A = \(\frac{1}{3}-\frac{1}{99}\)

2A = \(\frac{33}{99}-\frac{1}{99}\)

2A = \(\frac{32}{99}\)

A = \(\frac{32}{99}:2\)

A = \(\frac{32}{99}.\frac{1}{2}\)

A = \(\frac{32}{198}\)

A = \(\frac{16}{99}\)

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