9. $f$ is a function mapping positive integers to positive integers, $f(1)=5, f(m+n)=f(m)+$ $f(n)+5 m n-1\left(m, n \in \mathbf{N}_{+}\right), f(n)=$ $\qquad$ .
9. $\frac{5}{2} n^{2}+\frac{3}{2} n+1$.

Let $m=1$, then $f(n+1)=f(1)+f(n)+5 n-1=f(n)+5 n+4$, which means $f(n+1)-f(n)=5 n+4$. Therefore, $f(n+1)-f(1)=5(1+2+\cdots+n)+4 n=\frac{5}{2} n(n+1)+4 n$. So $f(n+1)=\frac{5}{2} n(n+1)+4 n+5, f(n)=\frac{5}{2} n(n-1)+4 n+1=\frac{5}{2} n^{2}+\frac{3}{2} n+1$