# mathematical induction

• Mar 20th 2008, 11:11 AM
Percent09
mathematical induction
Here is the problem:

Prove, by induction, that the sum (-1)^k (2k+1) from k=1 to 2n is proportional to n, and find the constant of proportionality.

Well I don't know how to start the induction part but I have a hunch that the constant of proportionality is just going to be 2.

Thank you
• Mar 20th 2008, 11:24 AM
topsquark
Quote:

Originally Posted by Percent09
Here is the problem:

Prove, by induction, that the sum (-1)^k (2k+1) from k=1 to 2n is proportional to n, and find the constant of proportionality.

Well I don't know how to start the induction part but I have a hunch that the constant of proportionality is just going to be 2.

Thank you

The base case is trivial (everything is proportional to 1), so let n = 2. Then
$\displaystyle \sum_{k = 1}^{4} (-1)^k (2k + 1) = -(2(1) + 1) + (2(2) + 1) - (2(3) + 1) + (2(4) + 1)$$\displaystyle = -3 + 5 - 7 + 9 = 4 \propto 2$

So assume that the theorem is true for some n = N. Then we need to show that it is true for n = N + 1. Let the constant of proportionality be 2, since this is true of the n = 2 case.
$\displaystyle \sum_{k = 1}^{2N + 2} (-1)^k (2k + 1) = \sum_{k = 1}^{2N} (-1)^k (2k + 1) - (2(N + 1) + 1) + (2(N + 2) + 1)$

$\displaystyle = 2N + 2 = 2(N + 1)$

This is proportional to N + 1 with a constant of proportionality 2. So etc, etc.

-Dan