Root, Root, Square Root

$\begin{aligned} S & = \sum_{n=1}^{24} \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} \\ & = \sum_{n=1}^{24} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{((n+1)\sqrt{n} + n\sqrt{n+1})((n+1)\sqrt{n} - n\sqrt{n+1}} \\ & = \sum_{n=1}^{24} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)^2 - n^2(n+1)} \\ & = \sum_{n=1}^{24} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)} \\ & = \sum_{n=1}^{24} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right) \\ & = \sum_{n=1}^{24} \frac{1}{\sqrt{n}} - \sum_{n=2}^{25} \frac{1}{\sqrt{n}} \\ & = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{25}} \\ & = \frac{\sqrt{25} -1} {\sqrt{25}} \end{aligned}$

$\Rightarrow S = \boxed{0.8}$

Summation is of the form: $\large \displaystyle \sum_{n=1}^m \dfrac{1}{\sqrt{n(n+1)}(\sqrt n+ \sqrt{n+1})}$ Rationalising it we get: $\large \displaystyle \sum_{n=1}^m \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}$ $\large =\color{#0C6AC7}{ \displaystyle \sum_{n=1}^m \dfrac{1}{\sqrt n}- \dfrac{1}{\sqrt {n+1}}}~~\small{\color{#20A900}{(Telescopic~series)}}$ $= 1- \dfrac{1}{\sqrt {m+1}}= \dfrac{\sqrt {m+1}-1}{\sqrt {m+1}}$ For this question $\color{#D61F06}{m=24 }$ , we get $\Large \dfrac{\sqrt{25}-1}{\sqrt{25}}=\boxed{\color{forestgreen}{\boxed{0.8}}}$