Difference between revisions of "1953 AHSME Problems/Problem 35"

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==Problem==
 
If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals:  
 
If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals:  
  
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\textbf{(C)}\ x(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}    </math>
 
\textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}    </math>
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==Solution==

Revision as of 20:19, 24 January 2020

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