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Question Number 123256 by ZiYangLee last updated on 24/Nov/20

$$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\dots +\frac{100}{98!+99!+100!}=?$$

Commented by ZiYangLee last updated on 24/Nov/20

$$answergiven:\frac{1}{2!}-\frac{1}{100!}$$

Commented by zarminaawan last updated on 24/Nov/20

$$howiuploadmyquestiosplzztellme?$$

Commented by Ar Brandon last updated on 24/Nov/20

Commented by Ar Brandon last updated on 24/Nov/20

While on forum tap there.

Answered by Olaf last updated on 24/Nov/20

$$\mathrm{S}=\underset{n=3}{\sum ^{100}}\frac{n}{(n-2)!+(n-1)(n-2)!+n(n-1)(n-2)!}$$$$\mathrm{S}=\underset{n=3}{\sum ^{100}}\frac{n}{(n-2)![1+(n-1)+n(n-1)]}$$$$\mathrm{S}=\underset{n=3}{\sum ^{100}}\frac{n}{(n-2)!n^2}=\underset{n=3}{\sum ^{100}}\frac{1}{(n-2)!n}$$$$\mathrm{S}=\underset{n=3}{\sum ^{100}}\frac{n-1}{n!}=\underset{n=3}{\sum ^{100}}(\frac{n}{n!}-\frac{1}{n!})=\underset{n=3}{\sum ^{100}}(\frac{1}{(n-1)!}-\frac{1}{n!})$$$$\mathrm{S}=\frac{1}{2!}-\frac{1}{100!}$$

Answered by MJS_new last updated on 24/Nov/20

$$\underset{k=m}{\sum ^n}\frac{k}{k!+(k-1)!+(k-2)!}=\frac{1}{(m-1)!}-\frac{1}{n!}$$$$(\mathrm{found}\mathrm{this}\mathrm{in}\mathrm{my}\mathrm{formulary},\mathrm{cannot}\mathrm{show}\mathrm{how}$$

Commented by ZiYangLee last updated on 25/Nov/20

$$\mathrm{oh}\mathrm{ok}\mathrm{thanks}\mathrm{sir}\mathrm{haha}$$

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