Find the value of : Find the value of $$\frac{\sqrt{100+\sqrt{1}}+\sqrt{100+\sqrt{2}}+\sqrt{100+\sqrt{3}}+\cdots+\sqrt{100+\sqrt{9999}}}{\sqrt{100-\sqrt{1}}+\sqrt{100-\sqrt{2}}+\sqrt{100-\sqrt{3}}+\cdots+\sqrt{100-\sqrt{9999}}}.$$

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.
