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1-1-2-1-2-3-1-19-20-1-20-21-




Question Number 163849 by alephzero last updated on 11/Jan/22
(1/(1 ∙ 2)) + (1/(2 ∙ 3)) + ... (1/(19 ∙ 20)) + (1/(20 ∙ 21)) = ?
$$\frac{\mathrm{1}}{\mathrm{1}\:\centerdot\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\:\centerdot\:\mathrm{3}}\:+\:…\:\frac{\mathrm{1}}{\mathrm{19}\:\centerdot\:\mathrm{20}}\:+\:\frac{\mathrm{1}}{\mathrm{20}\:\centerdot\:\mathrm{21}}\:=\:? \\ $$
Answered by cortano1 last updated on 11/Jan/22
 Σ_(k=1) ^(20)  (1/(k(k+1))) = Σ_(k=1) ^(20)  ((1/k)−(1/(k+1)))   = (1−(1/2))+((1/2)−(1/3))+...+((1/(20))−(1/(21)))   = 1−(1/(21)) = ((20)/(21))
$$\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\:\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$\:=\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{21}}\right) \\ $$$$\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{21}}\:=\:\frac{\mathrm{20}}{\mathrm{21}} \\ $$
Commented by alephzero last updated on 11/Jan/22
Thank You very much, sir!
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{very}\:\mathrm{much},\:\mathrm{sir}! \\ $$

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