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n-1-k-1-n-2-2n-




Question Number 191889 by cortano12 last updated on 03/May/23
      Σ_(n=1) ^k  (1/(n^2 +2n)) =?
$$\:\:\:\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\:=? \\ $$
Answered by mahdipoor last updated on 03/May/23
Σ_(n=1) ^k  (1/(n^2 +2n))=Σ_(n=1) ^k (1/2)((1/n)−(1/(n+2)))=  (1/2)[(1/1)+(1/2)−(1/(k+1))−(1/(k+2))]=(3/4)−((2k+3)/(2(k+1)(k+2)))
$$\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}{n}}=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{2}}\right]=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}{k}+\mathrm{3}}{\mathrm{2}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mehdee42 last updated on 03/May/23
      Σ_(n=1) ^k  (1/(n^2 +2n)) =(1/2)Σ_(n=1) ^k ((1/n)−(1/(n+2)))=(1/2)[(1−(1/3))+((1/2)−(1/4))+((1/3)−(1/5))+((1/4)−(1/5))+...+((1/(k−1))−(1/(k+1)))+((1/k)−(1/(k+2)))  =(1/2)((3/2)−((2k+3)/((k+1)(k+2))))=((k(3k+5))/(4(k+1)(k+2))) ✓
$$\:\:\:\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}\right)+…+\left(\frac{\mathrm{1}}{{k}−\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{2}}\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}{k}+\mathrm{3}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\right)=\frac{{k}\left(\mathrm{3}{k}+\mathrm{5}\right)}{\mathrm{4}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:\checkmark \\ $$$$ \\ $$

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