7. For any real numbers $x, y$, the function $f(x)$ satisfies
$$
f(x)+f(y)=f(x+y)-xy-1 \text{.}
$$

If $f(1)=1$, then for negative integers $n$, the expression for $f(n)$ is $\qquad$.
7. $f(n)=\frac{n^{2}+3 n-2}{2}$.

Let $y=1$, by the given condition we have
$$
f(x+1)=f(x)+f(1)+x+1=f(x)+x+2 \text {, }
$$

i.e., $f(x+1)-f(x)=x+2$.
In particular, $f(1)-f(0)=2$, which means $f(0)=-1$.
When $n$ is a negative integer,
$$
\begin{array}{l}
f(n)=[f(n)-f(n+1)]+[f(n+1)- \\
f(n+2)]+\cdots+[f(-1)-f(0)]+f(0) \\
=-[(n+2)+(n+3)+\cdots+1]-1 \\
=-\frac{(-n)(n+3)}{2}-1=\frac{n^{2}+3 n-2}{2} .
\end{array}
$$