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Question Number 163849 by alephzero last updated on 11/Jan/22

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...\frac{1}{19\cdot 20}+\frac{1}{20\cdot 21}=?$$

Answered by cortano1 last updated on 11/Jan/22

$$\underset{k=1}{\sum ^{20}}\frac{1}{k(k+1)}=\underset{k=1}{\sum ^{20}}(\frac{1}{k}-\frac{1}{k+1})$$$$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{20}-\frac{1}{21})$$$$=1-\frac{1}{21}=\frac{20}{21}$$

Commented by alephzero last updated on 11/Jan/22

$$\mathrm{Thank}\mathrm{You}\mathrm{very}\mathrm{much},\mathrm{sir}!$$

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