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Question Number 121246 by benjo_mathlover last updated on 06/Nov/20

$$\underset{\mathrm{k}=1}{\sum ^{49}}\frac{1}{\sqrt{\mathrm{k}+\sqrt{{\mathrm{k}}^2-1}}}?$$

Answered by liberty last updated on 06/Nov/20

$$\frac{1}{\sqrt{\mathrm{k}+\sqrt{{\mathrm{k}}^2-1}}}=\frac{1}{\sqrt{{(\sqrt{\frac{\mathrm{k}+1}{2}}+\sqrt{\frac{\mathrm{k}-1}{2}})}^2}}$$$$=\frac{1}{\sqrt{\frac{\mathrm{k}+1}{2}}-\sqrt{\frac{\mathrm{k}-1}{2}}}=\frac{\sqrt{\frac{\mathrm{k}+1}{2}}-\sqrt{\frac{\mathrm{k}-1}{2}}}{\frac{\mathrm{k}+1}{2}-\frac{\mathrm{k}-1}{2}}$$$$=\sqrt{\frac{\mathrm{k}+1}{2}}-\sqrt{\frac{\mathrm{k}-1}{2}}$$$$\mathrm{so}\underset{\mathrm{k}=1}{\sum ^{49}}(\sqrt{\frac{\mathrm{k}+1}{2}}-\sqrt{\frac{\mathrm{k}-1}{2}})=\sqrt{\frac{49+1}{2}}+\sqrt{\frac{48+1}{2}}-\sqrt{\frac{1}{2}}-0$$$$=5+\frac{7\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=5+3\sqrt{2}$$

Answered by Dwaipayan Shikari last updated on 06/Nov/20

$$\underset{k=1}{\sum ^{49}}\frac{1}{\sqrt{k+\sqrt{k^2-1}}}$$$$\underset{k=1}{\sum ^{49}}\sqrt{k-\sqrt{k^2-1}}$$$$\sqrt{k-\sqrt{k^2-1}}=\sqrt{\frac{k+\sqrt{k^2-(k^2-1)}}{2}}-\sqrt{\frac{k-\sqrt{k^2-(k^2-1)}}{2}}$$$$\underset{k=1}{\sum ^{49}}\sqrt{k-\sqrt{k^2-1}}=\sum ^{49}\sqrt{\frac{k+1}{2}}-\sqrt{\frac{k-1}{2}}$$$$=\sqrt{\frac{2}{2}}-\sqrt{\frac{0}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}+\sqrt{\frac{4}{2}}-\sqrt{\frac{2}{2}}+...\sqrt{\frac{50}{2}}-\sqrt{\frac{48}{2}}$$$$=-\sqrt{\frac{1}{2}}+\sqrt{\frac{50}{2}}+\sqrt{\frac{49}{2}}$$$$=\frac{7-1}{\sqrt{2}}+5=5+3\sqrt{2}$$

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