Fractional Summations 2018.

$\dfrac{\sum_{n=1}^{99} \sqrt{10 + \sqrt n}}{\sum_{n=1}^{99}\sqrt{10 - \sqrt n}} = \dfrac{\sum_{n=1}^{99} \sqrt{20 + 2\sqrt n}}{\sum_{n=1}^{99}\sqrt{20 - 2\sqrt n}}$

Note that $\sqrt{ 20+2\sqrt {n}} = \sqrt{10+ \sqrt{n}} + \sqrt{10- \sqrt{n}}$ Further note that $n$ and $100-n$ are symmteric for $1\leq N \leq 99$ . $\small\sqrt{20+\sqrt{n}} = \sqrt{10+ \sqrt{100-n}} + \sqrt{10- \sqrt{100-n}}$ therefore, $\dfrac{\sum_{n=1}^{99} \sqrt{20 + 2\sqrt n}}{\sum_{n=1}^{99}\sqrt{20 - 2\sqrt n}} = \dfrac{\sum_{n=1}^{99} \sqrt{10 + \sqrt {100-n}} + \sqrt{10 + \sqrt{100-n}}}{\sum_{n=1}^{99}\sqrt{10 - \sqrt {100-n}}- \sqrt{10 - \sqrt{100-n}}}$ Now let us assign $\sum_{n=1}^{99} \sqrt{10 + \sqrt n} = a$ and $\sum_{n=1}^{99} \sqrt{10 - \sqrt n} = b$ . Hence we can write the above summation as $\dfrac{a}{ b} = \dfrac{a+b}{a-b} \\ a^2 +2ab -b^2 =0 \\ \left(\dfrac{a}{b}\right)^2 + 2\left(\dfrac{a}{b}\right) -1 =0$ Solving quadratic equation in $\dfrac{a}{b}$ is $\dfrac{a}{b} = 1 + \sqrt 2$ . Therefore, the integral part(integer part) is $\boxed{2}$ .