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Question Number 80362 by Power last updated on 02/Feb/20

Answered by key of knowledge last updated on 02/Feb/20

$$\frac{1}{1+2+...+\mathrm{i}}-\frac{1}{1+2+...+\mathrm{i}+(\mathrm{i}+1)}=\frac{(1+...+\mathrm{i}+(\mathrm{i}+1))-(1+...+\mathrm{i})}{(1+2+...+\mathrm{i})(1+2+...+\mathrm{i}+(\mathrm{i}+1))}=$$$$\frac{(\mathrm{i}+1)}{(1+2+...+\mathrm{i})(1+2+...+\mathrm{i}+(\mathrm{i}+1))}$$$$\Rightarrow 1-[(\frac{1}{1}-\frac{1}{1+2})+(\frac{1}{1+2}-\frac{1}{1+2+3})+...+(\frac{1}{1+...+99}-\frac{1}{1+...+100})]=$$$$1-[1-\frac{1}{1+...+100}]=\frac{1}{5050}$$

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