Unformatted text preview: Recalling that any entry in Pascal's triangle is the sum of two entries directly above it, we conclude that the same holds here, but to diagonals instead of rows. The reader is invited to reflect on the identity [~] = [~::::~] + [nk"l] used in the solution to Exercise 24(c). Chapter 7 Review 1. For n = 1, P(2,1) = 2 is divisible by 2 1 , so the result is true. Assume that k ~ 1 and that 2k P(2k, k). We have P(2(k 1) k 1) = [2(k + 1)]! = (2k + 2)! + ,+ (k + 1)! (k + 1)! (2k + 2)(2k + 1)(2k)! (k + 1)! = 2(k + 1)(2k + 1)(2k)! = 2(2k + 1)(2k)! = 2(2k )P(2k k) (k + l)k! k! + 1 ,. Since P(2k, k) is divisible by 2k (by the induction hypothesis), we see that P(2(k + 1), k + 1) is divisible by 2 k + 1 , the desired result with n = k + 1. By the Principle of Mathematical Induction 2 n I P(2n, n) for all n ~ 1....
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 Summer '10
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 Binomial Theorem, Graph Theory, Mathematical Induction, OS, 2k

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