Three, (15 points) Find the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that
(1) $f(1)=1$;
(2) For all $x, y \in \mathbf{N}_{+}$,
$$
f(x+y)=f(x)+f(y)+x y
$$

holds.
Three, let the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ satisfy the given conditions.
For positive integers $n, k$ we have
$$
f((k+1) n)=f(k n)+f(n)+k n^{2} \text {. }
$$

Let $k=1,2, \cdots, m-1$ and sum up to get
$$
\begin{array}{l}
f(m n)=m f(n)+[1+2+\cdots+(m-1)] n^{2} \\
=m f(n)+\frac{m(m-1)}{2} n^{2},
\end{array}
$$

which holds for all positive integers $m, n$.
In particular, when $n=1$,
$$
f(m)=\frac{m(m+1)}{2} \text {. }
$$

Equation (1) defines the function $f$ on the set of positive integers.
Upon verification, $f(m)=\frac{m(m+1)}{2}$ is the unique solution to the problem.