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Question Number 131484 by bemath last updated on 05/Feb/21

$$\mathrm{Express}\underset{\mathrm{n}=1}{\sum ^{49}}\frac{1}{\sqrt{\mathrm{n}+\sqrt{{\mathrm{n}}^2-1}}}\mathrm{as}a+b\sqrt{2}$$$$\mathrm{for}\mathrm{some}\mathrm{integers}a\mathrm{and}b$$

Answered by benjo_mathlover last updated on 05/Feb/21

Answered by mr W last updated on 05/Feb/21

$$\frac{1}{\sqrt{n+\sqrt{n^2-1}}}$$$$=\sqrt{n-\sqrt{n^2-1}}$$$$=\sqrt{n-\sqrt{(n-1)(n+1)}}$$$$=\sqrt{\frac{n-1}{2}+\frac{n+1}{2}-2\sqrt{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)}}$$$$=\sqrt{{\left(\sqrt{\frac{n-1}{2}}\right)}^2+{\left(\sqrt{\frac{n+1}{2}}\right)}^2-2\sqrt{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)}}$$$$=\sqrt{{(\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}})}^2}$$$$=\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}$$$$\underset{n=1}{\sum ^{49}}(\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}})$$$$=\underset{n=2}{\sum ^{50}}\sqrt{\frac{k}{2}}-\underset{k=0}{\sum ^{48}}\sqrt{\frac{k}{2}}$$$$=(\underset{k=0}{\sum ^{48}}\sqrt{\frac{k}{2}}+\sqrt{\frac{49}{2}}+\sqrt{\frac{50}{2}}-\sqrt{\frac{1}{2}}-\sqrt{\frac{0}{2}})-\underset{k=0}{\sum ^{48}}\sqrt{\frac{k}{2}}$$$$=\sqrt{\frac{49}{2}}+\sqrt{\frac{50}{2}}-\sqrt{\frac{1}{2}}-\sqrt{\frac{0}{2}}$$$$=\frac{7}{\sqrt{2}}+5-\frac{1}{\sqrt{2}}$$$$=5+\frac{6}{\sqrt{2}}$$$$=5+3\sqrt{2}$$

Answered by Dwaipayan Shikari last updated on 05/Feb/21

$$\sqrt{n+\sqrt{n^2-1}}=\sqrt{\frac{n+\sqrt{n^2-n^2+1}}{2}}+\sqrt{\frac{n-\sqrt{n^2-n^2+1}}{2}}$$$$=\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}$$$$\underset{n=1}{\sum ^{49}}\frac{1}{\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}}=\frac{1}{\sqrt{2}}\underset{n=1}{\sum ^{49}}\sqrt{n+1}-\sqrt{n-1}$$$$=\frac{1}{\sqrt{2}}(\sqrt{2}-0+\sqrt{3}-1+\sqrt{4}-\sqrt{2}+....+\sqrt{49}-\sqrt{47}+\sqrt{50}-\sqrt{48})$$$$=\frac{1}{\sqrt{2}}(5\sqrt{2}+\sqrt{49}-1)=5+3\sqrt{2}$$

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