Problem: If f(x)=x(x−1)2f(x)=\dfrac{x(x-1)}{2}f(x)=2x(x−1), then f(x+2)f(x+2)f(x+2) equals:
Answer Choices:
A. f(x)+f(2)f(x)+f(2)f(x)+f(2)
B. (x+2)f(x)(x+2) f(x)(x+2)f(x)
C. x(x+2)f(x)x(x+2) f(x)x(x+2)f(x)
D. xf(x)x+2\dfrac{x f(x)}{x+2}x+2xf(x)
E. (x+2)f(x+1)x\dfrac{(x+2) f(x+1)}{x}x(x+2)f(x+1) Solution:
f(x+2)=(x+2)(x+2−1)/2 and f(x+1)=(x+1)(x+1−1)/2.\begin{aligned} & f(x+2)=(x+2)(x+2-1) / 2 \text { and } \\ & f(x+1)=(x+1)(x+1-1) / 2 . \end{aligned} f(x+2)=(x+2)(x+2−1)/2 and f(x+1)=(x+1)(x+1−1)/2.
So (x+1)/2=f(x+1)/x;∴f(x+2)=(x+2)f(x+1)/x(x+1) / 2=f(x+1) / x ; \quad \therefore f(x+2)=(x+2) f(x+1) / x(x+1)/2=f(x+1)/x;∴f(x+2)=(x+2)f(x+1)/x.