Problem: If f(x)=x(x+1)(x+2)(x+3)f(x)=x^{(x+1)}(x+2)^{(x+3)}f(x)=x(x+1)(x+2)(x+3) then f(0)+f(−1)+f(−2)+f(−3)=f(0)+f(-1)+f(-2)+f(-3)=f(0)+f(−1)+f(−2)+f(−3)=
Answer Choices:
A. −8/9-8 / 9−8/9
B. 000
C. 8/98 / 98/9
D. 111
E. 10/910 / 910/9
Solution:
Since 0z=00^{z}=00z=0 for any z>0,f(0)=f(−2)=0z>0, f(0)=f(-2)=0z>0,f(0)=f(−2)=0. Since (−1)0=1(-1)^{0}=1(−1)0=1, f(0)+f(−1)+f(−2)+f(−3)=(−1)0(1)2+(−3)−2(−1)0=1+1(−3)2=109f(0)+f(-1)+f(-2)+f(-3)=(-1)^{0}(1)^{2}+(-3)^{-2}(-1)^{0}=1+\dfrac{1}{(-3)^{2}}=\dfrac{10}{9}f(0)+f(−1)+f(−2)+f(−3)=(−1)0(1)2+(−3)−2(−1)0=1+(−3)21=910.