3. Given a function $f(n)$ defined on the set of positive integers satisfies the conditions:
(1) $f(m+n)=f(m)+f(n)+m n\left(m, n \in \mathbf{N}_{+}\right)$;
(2) $f(3)=6$.
Then $f(2011)=$ . $\qquad$
3.2023066.

In condition (1), let $n=1$ to get
$$
f(m+1)=f(m)+f(1)+m \text {. }
$$

Let $m=n=1$ to get
$$
f(2)=2 f(1)+1 \text {. }
$$

Let $m=2, n=1$, and use condition (2) to get
$$
6=f(3)=f(2)+f(1)+2 \text {. }
$$

From equations (3) and (2), we get
$$
f(1)=1, f(2)=3 \text {. }
$$

Substitute into equation (1) to get
$$
\begin{array}{l}
f(m+1)-f(m)=m+1 . \\
\text { Hence } f(2011)=\sum_{k=1}^{2010}[f(k+1)-f(k)]+f(1) \\
=\sum_{k=1}^{2010}(k+1)+1=\frac{2011 \times 2012}{2} \\
=2023066 .
\end{array}
$$