1968 AHSME Problems/Problem 5: Difference between revisions
m fixed typo |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 10: | Line 10: | ||
== 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 == | ||
Latest revision as of 17:11, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
