\(\frac{2}{1}\)\(\times\)3+\(\frac{2}{3}\)\(\times\)5+\(\frac{2}{5}\)\(\times\)7+...+\(\frac{2}{99}\)\(\times\)101
=(1-\(\frac{1}{3}\))+(\(\frac{1}{3}\)-\(\frac{1}{5}\))+(\(\frac{1}{5}\)-\(\frac{1}{7}\))+...+(\(\frac{1}{99}\)-\(\frac{1}{100}\))
=1-\(\frac{1}{100}\)
=\(\frac{100}{101}\)
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