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Question Number 100179 by bobhans last updated on 25/Jun/20

Answered by john santu last updated on 25/Jun/20

$$\mathrm{Tools}\sqrt{\mathrm{A}+\sqrt{\mathrm{B}}}=\sqrt{\frac{1}{2}(\mathrm{A}+\sqrt{{\mathrm{A}}^2-\mathrm{B}})}+\sqrt{\frac{1}{2}(\mathrm{A}-\sqrt{{\mathrm{A}}^2-\mathrm{B}})}$$$$\Rightarrow \sqrt{\mathrm{m}+\sqrt{{\mathrm{m}}^2-1}}=\sqrt{\frac{1}{2}(\mathrm{m}+1)}+\sqrt{\frac{1}{2}(\mathrm{m}-1)}$$$$\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}\frac{1}{\sqrt{\frac{1}{2}(\mathrm{m}+1)}+\sqrt{\frac{1}{2}(\mathrm{m}-1)}}$$$$=\sqrt{2}\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}\frac{1}{\sqrt{\mathrm{m}+1}+\sqrt{\mathrm{m}-1}}$$$$=\frac{1}{\sqrt{2}}\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}\sqrt{\mathrm{m}+1}-\sqrt{\mathrm{m}-1}$$$$=\frac{1}{\sqrt{2}}\underset{\mathrm{m}=3}{\sum ^{\mathrm{n}+2}}\sqrt{\mathrm{m}-1}-\frac{1}{\sqrt{2}}\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}\sqrt{\mathrm{m}-1}$$$$=\frac{1}{\sqrt{2}}\{\underset{\mathrm{m}=3}{\sum ^{\mathrm{n}}}\sqrt{\mathrm{m}-1}+\sqrt{\mathrm{n}}+\sqrt{\mathrm{n}+1}-(\sqrt{0}+\sqrt{1}+\underset{\mathrm{m}=3}{\sum ^{\mathrm{n}}}\sqrt{\mathrm{m}-1})\}$$$$=\frac{1}{\sqrt{2}}(\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}-1)◼$$

Answered by maths mind last updated on 25/Jun/20

$${(\sqrt{m+1}+\sqrt{m-1})}^2=2m+2\sqrt{m^2-1}$$$$\Rightarrow m+\sqrt{m^2-1}=\frac{{(\sqrt{m+1}+\sqrt{m-1})}^2}{2}$$$$=\sqrt{m+\sqrt{m^2-1}}=\frac{\sqrt{m+1}-\sqrt{m-1}}{\sqrt{2}}$$$$\mathrm{\Sigma }\frac{1}{\sqrt{m+\sqrt{m^2-1}}}=\underset{m=1}{\sum ^n}\frac{\sqrt{2}}{\sqrt{m+1}+\sqrt{m-1}}$$$$=\mathrm{\Sigma }\frac{\sqrt{2}(\sqrt{m+1}-\sqrt{m-1})}{2}=\frac{1}{\sqrt{2}}.\underset{m=1}{\sum ^n}(\sqrt{m+1}-\sqrt{m-1})$$$$=\frac{\sqrt{n+1}+\sqrt{n}-1}{\sqrt{2}}$$

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