The recursive function is
$f(n)= \begin{cases} 0, & \text{if x \le 0}.\\\\ f(n-1) + n, & \text{otherwise}. \end{cases}$
Assume \$$f \$$ is of the form \$$\frac{n(n+1)}{2} \$$; then
$f(0) = \frac{0(0+1)}{2} = 0$
$f(n) = \frac{n(n+1)}{2} = f(n-1) + n = \frac{[n-1]([n-1]+1)}{2} + n$