Take A Closer Look

First, let's start rationalizing the denominators. To do this, we can multiply the term $\frac{1}{\sqrt{100}+\sqrt{99}}$ by $\frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}}$ to get $\sqrt{100}-\sqrt{99}$ .

We can repeat this same process all the way down our sum, and end up with $\sqrt{100}-\sqrt{99}+\sqrt{99}-\sqrt{98}+\sqrt{98}-\sqrt{97}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}$

Conveniently, all of the terms except for $\sqrt{100}-\sqrt{1}$ cancel! This is called a Telescoping Sum .

Now, we can conclude that our sum is $k=\sqrt{100}-\sqrt{1}=10-1=9$ , so $k^{2}=\boxed{81}$

Same logic as Brandon Monsen 's, just different presentation.

$\begin{aligned} k & = \sum_{n=1}^{99} \frac{1}{\sqrt{n}+\sqrt{n+1}} \\ & = \sum_{n=1}^{99} \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})} \\ & = \sum_{n=1}^{99} \frac{\sqrt{n+1}-\sqrt{n}}{n+1-n} \\ & = \sum_{n=1}^{99} \left(\sqrt{n+1}-\sqrt{n} \right) \\ & = \sum_{n=2}^{100} \sqrt{n} - \sum_{n=1}^{99} \sqrt{n} \\ & = \sqrt{100} - \sqrt{1} \\ & = 9 \end{aligned}$

$\Rightarrow k^2 = 9^2 = \boxed{81}$

use summation notation