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Mathematical Induction II

This lecture covers Putnam Question B-1 from 1981, involving finding the limit of a summation with polynomials. The solution involves expressing the individual polynomial terms in terms of n, simplifying them, dividing by n^5, and taking the limit as n approaches infinity. The correct answer is -1. Additionally, the lecture includes examples of "Find the Flaw" proofs and a clever solution for finding a formula for a series.

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Mathematical Induction II

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

  2. 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: 1 + 3 + 5 + … + (2n – 1) = (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.

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

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