Difference between revisions of "1953 AHSME Problems/Problem 35"
| Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
| + | First simplify <math>f(x+2)</math>: | ||
| + | <cmath>f(x+2)=\frac{(x+2)(x+1)}{2}.</cmath> | ||
| + | Then simplify the answers and see which one matches. | ||
| + | |||
| + | <math>f(x)+f(2)=\frac{x(x-1)}{2} + 1,</math> | ||
| + | |||
| + | <math>(x+2)f(x)=\frac{(x+2)x(x-1)}{2},</math> | ||
| + | |||
| + | <math>x(x+2)f(x)=\frac{(x+2)x^2(x-1)}{2},</math> | ||
| + | |||
| + | <math>\frac{xf(x)}{x+2}=\frac{x^2(x-1)}{2(x+2)},</math> | ||
| + | |||
| + | <math>\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}.</math> | ||
| + | |||
| + | Since <math>f(x+2)=\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}</math>, the answer is <math>\boxed{\textbf{(E)}\ \frac{(x+2)f(x+1)}{x}}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 20:32, 24 January 2020
Problem
If 
, then 
equals:
Solution
First simplify :
![\[f(x+2)=\frac{(x+2)(x+1)}{2}.\]](http://latex.artofproblemsolving.com/c/e/4/ce4631bcb32d5ad2e5769a37a0c9c7cf3a29916b.png)
Then simplify the answers and see which one matches.
Since 
, the answer is 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
