In the problem I am not able to derive difference term. I know for solving summation we have to make difference term. How to proceed with this problem?
$$\frac{{\displaystyle \sum_{n=1}^{99}} \sqrt{10+\sqrt{n}}}{{\displaystyle \sum_{n=1}^{99}} \sqrt{10-\sqrt{n}}}$$
This can be done using an identity I learned from math.SE.
For any $a > 0$ and $0 \le b \le a^2$, we have $$\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}} = \sqrt{2}\sqrt{a + \sqrt{a^2-b}}$$
To prove this identity, just take squares on both sides and use the fact
$$\begin{align} \left(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\right)^2 &= \left(a + \sqrt{b}\right) + \left(a - \sqrt{b}\right) + 2\sqrt{a^2-b}\\ &= 2\left(a + \sqrt{a^2-b}\right)\end{align}$$
Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us
$$\sum_{b=1}^{a^2-1} \sqrt{a + \sqrt{b}} + \sum_{b=1}^{a^2-1}\sqrt{a - \sqrt{b}} = \sqrt{2}\sum_{b=1}^{a^2-1}\sqrt{a + \sqrt{a^2-b}} = \sqrt{2}\sum_{c=1}^{a^2-1}\sqrt{a + \sqrt{c}}$$ This leads to $$\frac{\sum_{b=1}^{a^2-1}\sqrt{a+\sqrt{b}}}{\sum_{b=1}^{a^2-1}\sqrt{a-\sqrt{b}}} = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1 $$ Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $\sqrt{2}+1$.
