2. $[x]$ represents the greatest integer not exceeding $x$, then
$$
\sum_{k=1}^{99}[\sqrt{k}]=
$$
$\qquad$
2. 615 .

Obviously, when $i^{2} \leqslant k \leqslant(i+1)^{2}-1=i^{2}+2 i(i=1,2,3$ $\cdots, 9)$, we have $[\sqrt{k}]=i$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{9}[\sqrt{k}]=\sum_{i=1}^{9} i(2 i+1) \\
=2 \cdot \sum_{i=1}^{9} i^{2}+\sum_{i=1}^{9} i \\
=2 \times \frac{9 \times 10 \times 19}{6}+\frac{9 \times 10}{2}=615 .
\end{array}
$$