Problem: Let f(n)=nx1+x2+…+xn. where n is a positive integer. If xk =(−1)k,k=1,2,3,…,n, the set of possible values of f(n) is:
Answer Choices:
A. {0}
B. {n1}
C. {0,−n1}
D. {0,n1}
E. {1,n1}
Solution:
Since xk=(−1)k,x1=−1,x2=1,…,x2r−1=−1,x2r=1. Therefore, for n=2r, x1+x2+…+x2r=0 and for n=2r−1,x1+x2+…+x2r−1=−1. Therefore, f(n)=0 in the former case and f(n)=−n1 in the latter case.