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Question Number 112881 by bemath last updated on 10/Sep/20

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...$$$$\underset{n=1}{\sum ^n}\frac{2}{n(n+1)}=2\underset{n=1}{\sum ^{40}}\frac{1}{n}-\frac{1}{n+1}=2(1-\frac{1}{41})=\frac{80}{41}$$

Commented by bemath last updated on 10/Sep/20

$$\mathrm{sir}\mathrm{how}\mathrm{you}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{formula}\frac{2}{\mathrm{n}(\mathrm{n}+1)}$$

Commented by Dwaipayan Shikari last updated on 10/Sep/20

$$Termnth$$$$=\frac{1}{1+2+3+4+5+6+7+...+n}=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$$$$$$

Commented by bemath last updated on 10/Sep/20

$$\mathrm{oo}\mathrm{thank}\mathrm{you}$$

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