$7 \cdot 112$ If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals
(A) $f(x)+f(2)$.
(B) $(x+2) f(x)$.
(C) $x(x+2) f(x)$.
(D) $\frac{x f(x)}{x+2}$.
(E) $\frac{(x+2) f(x+1)}{x}$.
(4th American High School Mathematics Examination, 1953)
[Solution] Given $f(x+2)=\frac{(x+2)[(x+2)-1]}{2}$

and
$$
=\frac{(x+1)(x+2)}{2} \text {, }
$$

thus
$$
\begin{aligned}
f(x+1) & =\frac{(x+1)[(x+1)-1]}{2} \\
& =\frac{x(x+1)}{2}
\end{aligned}
$$

Therefore, the answer is $(E)$.