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Question Number 22062 by x² - y²@gmail.com last updated on 10/Oct/17

Commented by FilupS last updated on 10/Oct/17

$$=\underset{n=1}{\sum ^{97}}\frac{1}{(n+1)!\cdot (n+3)!}$$$$=\underset{n=1}{\sum ^{97}}\frac{1}{(n+1){!}^2\cdot (n+2)(n+3)}$$$$=???$$

Answered by $@ty@m last updated on 11/Oct/17

$$=(\frac{1}{3!}-\frac{1}{4!})+(\frac{1}{4!}-\frac{1}{5!})+(\frac{1}{5!}-\frac{1}{6!})+...+(\frac{1}{99!}-\frac{1}{100!})$$$$=\frac{1}{3!}-\frac{1}{100!}$$

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