Difference between revisions of "1968 AHSME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | |
| + | Plugging in the expressions for <math>f(r)</math> and <math>f(r-1)</math>, we see that: | ||
| + | \begin{align*} | ||
| + | f(r)-f(r-1) | ||
| + | &= \frac{1}{3}r(r+1)(r+2)-\frac{1}{3}(r-1)(r-1+1)(r-1+2) \\ | ||
| + | &=\frac{1}{3}[r(r+1)(r+2)-r(r-1)(r+1)] \\ | ||
| + | &=\frac{1}{3}[r(r^2+3r+2)-r(r^2-1)] \\ | ||
| + | &=\frac{1}{3}[r^3+3r^2+2r-r^3+r] \\ | ||
| + | &=\frac{1}{3}[3r^2+3r] \\ | ||
| + | &=r^2+r \\ | ||
| + | &=r(r+1), \\ | ||
| + | \end{align*} | ||
| + | which is answer choice <math>\fbox{A}</math> | ||
== See also == | == See also == | ||
Revision as of 17:07, 17 July 2024
Problem
If 
, then 
equals:
Solution
Plugging in the expressions for 
and 
, we see that:
\begin{align*}
f(r)-f(r-1)
&= \frac{1}{3}r(r+1)(r+2)-\frac{1}{3}(r-1)(r-1+1)(r-1+2) \\
&=\frac{1}{3}[r(r+1)(r+2)-r(r-1)(r+1)] \\
&=\frac{1}{3}[r(r^2+3r+2)-r(r^2-1)] \\
&=\frac{1}{3}[r^3+3r^2+2r-r^3+r] \\
&=\frac{1}{3}[3r^2+3r] \\
&=r^2+r \\
&=r(r+1), \\
\end{align*}
which is answer choice 
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
