3. Let $a_{n}=\sqrt{1 \times 2}+\sqrt{2 \times 3}+\cdots+\sqrt{n \times(n+1)}$. Then $\left[\frac{2 a_{n}}{n}\right]=$ $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$).
3. $n+1$.

Since $k<\sqrt{k(k+1)}<k+\frac{1}{2}$, we have
$$
\sum_{k=1}^{n} k<a_{n}<\sum_{k=1}^{n}\left(k+\frac{1}{2}\right) \text {, }
$$

which means $\frac{n(n+1)}{2}<a_{n}<\frac{n(n+2)}{2}$.
Therefore, $\left[\frac{2 a_{n}}{n}\right]=n+1$.