1968 AHSME Problems/Problem 5

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: \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 $\fbox{A}$

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1968 AHSME Problems/Problem 5

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