Simplify $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots+ \frac{1}{\sqrt{24} + \sqrt{25}}$ Simplify$$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}.$$
I know you can solve this using generating functions but I'm not totally sure.

Here, we simply use André Nicolas' hint to observe the "collapsing house of cards":
We first represent the $k$th term of the sum, to simplify matters.
$$\sum_{k = 1}^{24} \color{blue}{\bf \frac{1}{\sqrt k + \sqrt{k+1}}}\quad = \quad\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}$$
$$ {\bf NOTE:}\quad \frac{1}{\sqrt k + \sqrt{k+1}}\cdot \frac{\sqrt k - \sqrt{k+1}}{\sqrt k - \sqrt{k+1}} =  \frac{\sqrt k - \sqrt{k+1}}{k - (k + 1)} = \color{blue}{\bf \sqrt {k+1} - \sqrt{k}}$$
$$ $$
$$
\begin{align}\sum_{k = 1}^{24} \frac{1}{\sqrt k + \sqrt{k+1}} \quad
& = \quad\sum_{k=1}^{24} \sqrt {k+1} - \sqrt{k} \\ \\
& = (\sqrt 2 - 1) + (\sqrt 3 - \sqrt 2) + (\sqrt 4 - \sqrt 3) \cdots + (\sqrt{25} -\sqrt {24}) \\ \\
& = \sqrt{25} - 1= 4 \\
\end{align}
$$
