2544.11 – Function as Average

Tags:Problem Set 3Pre-Calculus MathSymbol-Pushing

Let xk=(−1)kx_k = (-1)^k for any positive integer kk. Let f(n)=(x1+x2+...+xn)nf(n) = \frac{(x_1 + x_2 + ...+ x_n)}{n}, where nn is a positive integer.

Give the range of this function.

  1. 0
  2. 1/nn (where n is any positive integer)
  3. 0 and -1/nn (where n is any odd positive integer)
  4. 0 and 1/nn (where n is any positive integer)
  5. 1 and 1/nn (where n is any odd positive integer).
Solution

Let's compute:

xk=(−1)kx1=−1x2=1x3=−1x4=1. \begin{aligned} x_k &=& (-1)^k \\ x_1 &=& -1 \\ x_2 &=& 1 \\ x_3 &=& -1 \\ x_4 &=& 1. \end{aligned}

So, xn=−1x_n = -1 if nn is odd and xn=1x_n = 1 if nn is even.

Thus x1+x2+...+xnx_1 + x_2 + ... + x_n = 0 if nn is even, −1-1 if n is odd.

So,

f(1)=−11f(1) = \frac{-1}{1}= -1

f(2)=01f(2) = \frac{0}{1}= 0

f(3)=−13f(3) = \frac{-1}{3}

f(4)=04f(4) = \frac{0}{4}= 0

f(5)=−15f(5) = \frac{-1}{5}

etc

The answer is (c).