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Question Number 144981 by mathdanisur last updated on 01/Jul/21

$$\underset{k=1}{\sum ^{35}}\frac{\sqrt{k}}{k+\sqrt{k^2+k}}=?$$

Answered by liberty last updated on 01/Jul/21

$$\frac{\sqrt{\mathrm{k}}}{\mathrm{k}+\sqrt{{\mathrm{k}}^2+\mathrm{k}}}=\frac{\sqrt{\mathrm{k}}(\mathrm{k}-\sqrt{{\mathrm{k}}^2+\mathrm{k}})}{-\mathrm{k}}$$$$=-\sqrt{\mathrm{k}}+\frac{\sqrt{{\mathrm{k}}^2+\mathrm{k}}}{\sqrt{\mathrm{k}}}$$$$=-\sqrt{\mathrm{k}}+\sqrt{\mathrm{k}+1}$$$$=\sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}}$$$$\underset{\mathrm{k}=1}{\sum ^{35}}(\sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}})=\underset{\mathrm{telescopic}\mathrm{series}}{\underbrace{6-1=5}}$$

Commented by mathdanisur last updated on 01/Jul/21

$$thankyouSir,cool$$

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