Discrete Mathematics with Graph Theory (3rd Edition) 213

Discrete Mathematics with Graph Theory (3rd Edition) 213 -...

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Chapter 7 211 Since ( n-k+1) = (n-(n 2 1 +1)+1) = (~) =1 k n-1 + 1 .!tl.! 2 2 also, we again conclude that [~] = (n-Z+1). By the Principle of Mathematical Induction, we have the result. 25. As suggested, we create a triangle like that in Fig. 7.8, replacing (~) with [~] and obtain the follow- ing. 1 1 1 1 2 0 1 3 1 0 1 4 3 0 0 1 5 6 1 0 0 1 6 10 4 0 0 0 1 7 15 10 1 0 0 0 As proved in Exercise 24(a) , the left-most entry in each row is 1. Also, the result in (b) gives a substantial number of Os at the end of each row. The result in (c) is more interesting. It tells us, for example, that [~], m, [~], m, ... are just the binomial coefficients G), G), G), G), . ... But [~] = 1, m = 5, [~] = 10, m = 10 are also the elements on one of the downward left to right diagonals. In this way, we see that the present triangle is just a skewed version of Pascal's (with diagonals replacing rows), padded with Os.
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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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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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