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Question Number 191889 by cortano12 last updated on 03/May/23

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}\frac{1}{{\mathrm{n}}^2+2\mathrm{n}}=?$$

Answered by mahdipoor last updated on 03/May/23

$$\underset{n=1}{\sum ^k}\frac{1}{n^2+2n}=\underset{n=1}{\sum ^k}\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})=$$$$\frac{1}{2}[\frac{1}{1}+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}]=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}$$$$$$$$$$$$$$

Answered by mehdee42 last updated on 03/May/23

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}\frac{1}{{\mathrm{n}}^2+2\mathrm{n}}=\frac{1}{2}\underset{n=1}{\sum ^k}(\frac{1}{n}-\frac{1}{n+2})=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{5})+...+(\frac{1}{k-1}-\frac{1}{k+1})+(\frac{1}{k}-\frac{1}{k+2})$$$$=\frac{1}{2}(\frac{3}{2}-\frac{2k+3}{(k+1)(k+2)})=\frac{k(3k+5)}{4(k+1)(k+2)}✓$$$$$$

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