Let xk=(−1)k for any positive integer k. Let f(n)=n(x1+x2+...+xn), where n is a positive integer.
Give the range of this function.
- 0
- 1/n (where n is any positive integer)
- 0 and -1/n (where n is any odd positive integer)
- 0 and 1/n (where n is any positive integer)
- 1 and 1/n (where n is any odd positive integer).
Solution
Let's compute:
xkx1x2x3x4=====(−1)k−11−11.
So, xn=−1 if n is odd and xn=1 if n is even.
Thus x1+x2+...+xn = 0 if n is even, −1 if n is odd.
So,
f(1)=1−1= -1
f(2)=10= 0
f(3)=3−1
f(4)=40= 0
f(5)=5−1
etc
The answer is (c).