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.