Task B-4.2. The general term of the sequence $a_{n}$ is given by the formula $a_{n}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}$. Determine the sum of the first 9999 terms of this sequence.
## Solution.

The general term of the sequence can be written in the following form

$$
\begin{aligned}
a_{n} & =\frac{1}{(\sqrt{n+1})^{2} \sqrt{n}+(\sqrt{n})^{2} \sqrt{n+1}}=\frac{1}{\sqrt{n+1} \sqrt{n}(\sqrt{n+1}+\sqrt{n})}= \\
& =\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n} \sqrt{n+1}(n+1-n)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
\end{aligned}
$$

The desired sum is then

$a_{1}+a_{2}+\ldots+a_{9999}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{9999}}-\frac{1}{\sqrt{10000}}=1-\frac{1}{\sqrt{10000}}=\frac{99}{100}$.