7. (1976 Yugoslav Mathematical Olympiad) Let $a_{n}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\left(n \in \mathbf{N}^{*}\right)$, then
$$
\sum_{k=1}^{99} a_{k}=
$$
7. Slightly explained: Since $a_{n}=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1)^{2} n-n^{2}(n+1)}=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)}=\frac{1}{\sqrt{n}}-$ $\frac{1}{\sqrt{n+1}}$. Therefore, $\sum_{k=1}^{99} a_{k}=\sum_{k=1}^{99}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=\frac{9}{10}$.