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

2. 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.

3. 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.

4. 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.

5. 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.

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

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

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