Calculate $\frac{\sum_{n=1}^{99} \sqrt{10+\sqrt{n}}}{\sum_{n=1}^{99} \sqrt{10-\sqrt{n}}}$

Asked Dec 08 '19 at 20:33

Active Dec 08 '19 at 20:33

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I have to solve this question for a math class I'm taking, but I'm not making much headway. I tried to let $\sqrt{10+\sqrt{n}}=\sqrt{a}+\sqrt{b}$ in an attempt to make things cancel out, and got that:

$a+b=10$

$4ab=n$

But I don't think it helps me much.

asked Dec 08 '19 at 20:33

Silverleaf1

2

and in general it seems $\dfrac{\sum\limits_{n=1}^{m^2-1} \sqrt{m+\sqrt{n}}}{\sum\limits_{n=1}^{m^2 -1} \sqrt{m-\sqrt{n}}} = 1+\sqrt 2$ – Henry Dec 08 '19 at 20:43
5

Does this answer your question? Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ – conditionalMethod Dec 08 '19 at 20:46
@conditionalMethod - it seems I made the same comment there too – Henry Dec 08 '19 at 20:48
Some other approaches using trigonometry are here. – conditionalMethod Dec 08 '19 at 20:50
While this uses $\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=\sqrt2+1$ – Henry Dec 08 '19 at 20:53
See also my answer here. – Ng Chung Tak Dec 27 '19 at 11:41

Calculate $\frac{\sum_{n=1}^{99} \sqrt{10+\sqrt{n}}}{\sum_{n=1}^{99} \sqrt{10-\sqrt{n}}}$

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