Mathematical Induction II Lecture 20 Section 4.3 Wed, Feb 16, 2005

limn [(1/n5) h = 1..n k = 1..n (5h4 – 18h2k2 + 5k4)]. Putnam Question B-1 (1981) Find limn [(1/n5) h = 1..n k = 1..n (5h4 – 18h2k2 + 5k4)]. Solution Express hk 5h4, hk 18h2k2, and hk 5k4 as polynomials in n. Simplify the polynomials. Divide by n5. Take the limit as n  . The answer is -1.

Let’s Play “Find the Flaw” Theorem: For every positive integer n, in any set of n horses, all the horses are the same color. Proof: Basic Step. When n = 1, there is only one horse, so trivially they are (it is) all the same color.

Find the Flaw Inductive Step Suppose that any set of k horses are all the same color. Consider a set of k + 1 horses. Remove one of the horses from the set. The remaining set of k horses are all the same color.

Find the Flaw Replace that horse and remove a different horse. Again, the remaining set of k horses are all the same color. Therefore, the two horses that were removed are the same color as the other horses in the set. Thus, the k + 1 horses are all the same color.

Find the Flaw Thus, in any set of n horses, the horses are all the same color.

Example Find a formula for 1 + 3 + 5 + … + (2n – 1). Clever solution: = (1 + 2 + 3 + … + 2n) – (2 + 4 + 6 + … + 2n) = (1 + 2 + 3 + … + 2n) – 2(1 + 2 + 3 + … + n) =(2n)(2n + 1)/2 – 2(n(n + 1)/2) = n2.

Exercise Find a formula for 12 + 32 + 52 + … + (2n – 1)2. Then verify it using mathematical induction.