Find the value of : Find the value of $$\frac{\sqrt{100+\sqrt{1}}+\sqrt{100+\sqrt{2}}+\sqrt{100+\sqrt{3}}+\cdots+\sqrt{100+\sqrt{9999}}}{\sqrt{100-\sqrt{1}}+\sqrt{100-\sqrt{2}}+\sqrt{100-\sqrt{3}}+\cdots+\sqrt{100-\sqrt{9999}}}.$$
 A: It is curious that this is listed as "precalculus", because I'm not sure that a solid answer can be given for an exact value of this ratio. If you're allowed some calculus, then it's easy to see that the ratio can be written as
$$
 S = \frac{\sum_{n=1}^{9999}{\sqrt{1+\sqrt{n/10000}}}}{\sum_{n=1}^{9999}{\sqrt{1-\sqrt{n/10000}}}}
$$
From this, we can see that it's a quite high-resolution approximation to the ratio of integrals
$$
 S \approx \frac{\int_0^1 \sqrt{1+\sqrt{x}}dx}{\int_0^1 \sqrt{1-\sqrt{x}}dx} = \frac{\frac{8}{15}(\sqrt{2}+1)}{\frac{8}{15}} = \sqrt{2}+1
$$
And this is a good approximation to the ratio. I cannot see a way to get a simple algebraic value for it without calculus.
A: If you can proof for a fraction of the form
$\frac{\sum_i a_i}{\sum_i b_i}$
that $a_i=r b_i$ for all $i$ then obviously 
$$\frac{\sum_i a_i}{\sum_i b_i}=r$$
Hence, in the fraction given in the problem statement it suffices to show that
$$\frac{\sqrt{100+\sqrt{5000-i}}+\sqrt{100+\sqrt{5000+i}}}{\sqrt{100-\sqrt{5000-i}}+\sqrt{100-\sqrt{5000+i}}}=\sqrt{2}+1.$$
(Just sum the terms in the numerator and the denominator from $i=1$ to $4999$ and divide by $2$). To show the last equation set $100=\sqrt{2x}$, hence $5000=x$. Then it needs to be shown that
$$\frac{\sqrt{\sqrt{2x}+\sqrt{x-i}}+\sqrt{\sqrt{2x}+\sqrt{x+i}}}{\sqrt{\sqrt{2x}-\sqrt{x-i}}+\sqrt{\sqrt{2x}-\sqrt{x+i}}}=\sqrt{2}+1.$$
For the proof see the answer to this question (thanks Boris Novikov). Thus the value of the expression is $\sqrt{2}+1$.
