Problem: If f(n)=13n(n+1)(n+2)f(n)=\dfrac{1}{3} n(n+1)(n+2)f(n)=31n(n+1)(n+2), then f(r)−f(r−1)f(r)-f(r-1)f(r)−f(r−1) equals:
Answer Choices:
A. r(r+1)\mathrm{r}(\mathrm{r}+1)r(r+1)
B. (r+1)(r+2)(r+1)(r+2)(r+1)(r+2)
C. 13r(r+1)\dfrac{1}{3} r(r+1)31r(r+1)
D. 13(r+1)(r+2)\dfrac{1}{3}(r+1)(r+2)31(r+1)(r+2)
E. 13r(r+1)(2r+1)\dfrac{1}{3} r(r+1)(2 r+1)31r(r+1)(2r+1) Solution:
f(r)−f(r−1)=13r(r+1)(r+2)−13(r−1)(r)(r+1)=13r(r+1)(r+2−r+1)f(r)-f(r-1)=\dfrac{1}{3} r(r+1)(r+2)-\dfrac{1}{3}(r-1)(r)(r+1)=\dfrac{1}{3} r(r+1)(r+2-r+1)f(r)−f(r−1)=31r(r+1)(r+2)−31(r−1)(r)(r+1)=31r(r+1)(r+2−r+1)
=r(r+1).=r(r+1) .=r(r+1).