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Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 9 (due February 18) 7.1 # 5 r 2 8 r + 16 = ( r 4) 2 , so the general solution is c 1 4 n + c 2 n 4 n . a. c 1 = c 2 = 0, yes. b. No. c. No. d. Yes. e. Yes. f. Yes. g. No. h. No. 7.1 # 8 a. a n = ( 1) a n 1 = ( 1) 2 a n 2 = = ( 1) n a = ( 1) n 5 b. a n = a n 1 + 3 = a n 2 + 2(3) = = a + n (3) = 3 n + 1 c. a n = a n 1 n = a n 2 ( n + ( n 1)) = = a ( n + ( n 1) + + 2 + 1) = 4 n ( n +1) 2 . d. a n = 2 a n 1 3 = 2 2 a n 2 (6+3) = = 2 n a (3 + 6 + + 3 2 n 1 ) = 1 3 (2 n 1) = 2 3 2 n e. a n = ( n +1) a n 1 = ( n +1)( n ) a n 2 = ... ( n +1)( n )( n 1) ... 3 2 a = 2 ( n + 1)! f. a n = 2 na n 1 = 2 2 n ( n 1) a n 2 = ... 2 n n ( n 1) 2 1 a = 3 2 n n ! g. a n = a n 1 + n 1 = ( 1) n a n 2 + ( n 1) ( n 2) = = ( 1) n a + ( n 1) ( n 2) + ( n 3) ( n 4) = ( 1) n 7 + n 2 1 7.1 # 17 Base case n = 1: 0 = H = 2 1 = 1 1 = 0. Now assume for induction that H k = 2 k 1. Then H k +1 = 2 H k + 1 = 2 ( 2 k 1 ) + 1 = 2 k +1 1, which is what we needed to show. So by induction the formula is correct for all n . 7.1 # 20 There is no way to pay negative pesos, so x n = 0 for n < 0. There is one way to pay zero pesos (no coins), so x = 1. For n > 0, we just subtract the first coin. So: x n = x n 1 + x n 2 + 2 x n 5 + 2 x n 10 + x n 20 + x n 50 + x n 100 7.1 # 28 a. s n = s n 1 + s n 2 + s n 3 b. s = 1,...
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This note was uploaded on 07/17/2009 for the course MAT 55 taught by Professor Andrzejczygrinow during the Spring '09 term at University of California, Berkeley.
 Spring '09
 ANDRZEJCZYGRINOW
 Math

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