Question Number 22062 by x² – y²@gmail.com last updated on 10/Oct/17
Commented by FilupS last updated on 10/Oct/17
$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{97}} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!\centerdot\left({n}+\mathrm{3}\right)!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{97}} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!^{\mathrm{2}} \centerdot\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=??? \\ $$
Answered by $@ty@m last updated on 11/Oct/17
$$=\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}!}−\frac{\mathrm{1}}{\mathrm{6}!}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{99}!}−\frac{\mathrm{1}}{\mathrm{100}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{100}!} \\ $$