Difference between revisions of "1968 AHSME Problems/Problem 5"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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Plugging in the expressions for <math>f(r)</math> and <math>f(r-1)</math>, we see that:
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\begin{align*}
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f(r)-f(r-1)
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&= \frac{1}{3}r(r+1)(r+2)-\frac{1}{3}(r-1)(r-1+1)(r-1+2) \
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&=\frac{1}{3}[r(r+1)(r+2)-r(r-1)(r+1)] \
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&=\frac{1}{3}[r(r^2+3r+2)-r(r^2-1)] \
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&=\frac{1}{3}[r^3+3r^2+2r-r^3+r] \
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&=\frac{1}{3}[3r^2+3r] \
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&=r^2+r \
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&=r(r+1), \
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\end{align*}
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which is answer choice <math>\fbox{A}</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:07, 17 July 2024

Problem

If $f(n)=\tfrac{1}{3} n(n+1)(n+2)$, then $f(r)-f(r-1)$ equals:

$\text{(A) } r(r+1)\quad \text{(B) } (r+1)(r+2)\quad \text{(C) } \tfrac{1}{3} r(r+1)\quad  \\ \text{(D) } \tfrac{1}{3} (r+1)(r+2)\quad \text{(E )} \tfrac{1}{3} r(r+1)(2r+1)$

Solution

Plugging in the expressions for $f(r)$ and $f(r-1)$, we see that: f(r)f(r1)=13r(r+1)(r+2)13(r1)(r1+1)(r1+2)=13[r(r+1)(r+2)r(r1)(r+1)]=13[r(r2+3r+2)r(r21)]=13[r3+3r2+2rr3+r]=13[3r2+3r]=r2+r=r(r+1), which is answer choice $\fbox{A}$

See also

1968 AHSC (ProblemsAnswer KeyResources)
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

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