# 1953 AHSME Problems/Problem 35

## Problem

If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals:

$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$

## Solution

First simplify $f(x+2)$: $$f(x+2)=\frac{(x+2)(x+1)}{2}.$$ Then simplify the answers and see which one matches.

$f(x)+f(2)=\frac{x(x-1)}{2} + 1,$

$(x+2)f(x)=\frac{(x+2)x(x-1)}{2},$

$x(x+2)f(x)=\frac{(x+2)x^2(x-1)}{2},$

$\frac{xf(x)}{x+2}=\frac{x^2(x-1)}{2(x+2)},$

$\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}.$

Since $f(x+2)=\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}$, the answer is $\boxed{\textbf{(E)}\ \frac{(x+2)f(x+1)}{x}}.$

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 34 Followed byProblem 36 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions