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Question Number 207845 by hardmath last updated on 28/May/24

$$\mathrm{Find}:$$$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+40}$$

Answered by Frix last updated on 28/May/24

$$\underset{j=1}{\sum ^n}\left(\underset{k=1}{\sum ^j}k\right)=2-\frac{2}{n+1}$$$$n=40\Rightarrow \mathrm{answer}\mathrm{is}\frac{80}{41}$$

Commented by hardmath last updated on 28/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{professor}$$

Answered by MM42 last updated on 28/May/24

$$=\frac{2}{1\times 2}+\frac{2}{2\times 3}+\frac{2}{3\times 4}+...+\frac{2}{39\times 40}+\frac{2}{40\times 41}$$$$=2[(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{39}-\frac{1}{40})+(\frac{1}{40}-\frac{1}{41})]$$$$=2(1-\frac{1}{41})=\frac{80}{41}✓$$$$$$

Commented by hardmath last updated on 28/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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