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 by Problem 34	Followed by Problem 36
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1953 AHSME Problems/Problem 35

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