16. Suppose that the graph of quadratic function $f(x)$ passes through the origin, and there holds $f(x+2)=f(x)+x+2$ for all $x \in \mathbf{R}$. Then the expression of $f(x)$ is . $\qquad$
16. $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x$.

The problem states: If the graph of the quadratic function $f(x)$ passes through the origin, and for any $x \in \mathbf{R}$, $f(x+2)=f(x)+x+2$ holds, then the expression for $f(x)$ is $\qquad$ .
From the fact that the graph of $f(x)$ passes through the origin, we know $f(0)=0$.
Let $f(x)=a x^{2}+b x$. From
$f(x+2)=a(x+2)^{2}+b(x+2)=a x^{2}+b x+x+2$
we get $\left\{\begin{array}{l}4 a-1=0, \\ 4 a+2 b=2 .\end{array}\right.$
Solving these, we find $a=\frac{1}{4}, b=\frac{1}{2}$.
Therefore, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x$.