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Unformatted text preview: MATH 135 Fall 2005 Assignment 7 Solutions HandIn Problems Exercise 428: A sequence of integers x 1 , x 2 , x 3 , . . . is defined by x 1 = 3, x 2 = 7, and x k = 5 x k 1 6 x k 2 for k 3 . Prove that x n = 2 n + 3 n 1 for all n P . Solution: Base Cases When n = 1, x 1 = 3 = 2 1 + 3 so the result is true. When n = 2, x 2 = 7 = 2 2 + 3 1 so the result is true. Induction Hypothesis Assume that x r = 2 r + 3 r 1 for all 1 r k , for some k Z with k 2. Induction Conclusion Consider n = k + 1. Then x k +1 = 5 x k 6 x k 1 = 5(2 k + 3 k 1 ) 6(2 k 1 + 3 k 2 ) (by Induction Hypothesis) = 5 2 k + 5 3 k 1 3 2 k 2 3 k 1 = 2 2 k + 3 3 k 1 = 2 k +1 + 3 k . Therefore, the result holds for n = k + 1. Therefore, by the Principle of Strong Mathematical Induction x n = 2 n + 3 n 1 for all n P . Exercise 430: Find an expression for 1 3 + 5 7 + 9 11 + + ( 1) n 1 (2 n 1) and prove that it is correct. Solution: We first calculate the first few sums to try to determine the pattern: 1 = 1 1 3 = 2 1 3 + 5 = 3 1 3 + 5 7 = 4 By observing the above pattern, we can guess the following formula: 1 3 + 5 7 + + ( 1) n 1 (2 n 1) = ( 1) n 1 n. We shall prove that this result is true for all n P , by induction on n. Base Case If n = 1, the left side equals 1 and the right side equals ( 1) 1 1 (1) = 1, so the result is true. Induction Hypothesis Assume that 1 3 + 5 7 + + ( 1) k 1 (2 k 1) = ( 1) k 1 k. for some k P , k 1. Induction Conclusion Consider n = k + 1. Then 1 3 + 5 7 + + ( 1) k 1 (2 k 1) + ( 1) k (2( k + 1) 1) = ( 1) k 1 k + ( 1) k (2 k + 1) (by Induction Hypothesis) = ( 1) k ( k ) + ( 1) k (2 k + 1) = ( 1) k ( k + 2 k + 1) = ( 1) k ( k + 1) . Therefore the result holds for n = k + 1 . Therefore, by the Principle of Mathematical Induction, 1 3 + 5 7 + + ( 1) n 1 (2 n 1) = ( 1) n 1 n for all n P . Problem 460: Prove that gcd( f n , f n +1 ) = 1 for all n P . Solution: We shall prove the result by induction on n . Base Case When n = 1, gcd( f 1 , f 2 ) = gcd(1 , 1) = 1, so the result holds. Induction Hypothesis Assume that gcd( f k , f k +1 ) = 1 for some k P , k 1. Induction Conclusion Consider n = k + 1. We want to show that gcd( f k +1 , f k +2 ) = 1. Let d = gcd( f k +1 , f k +2 ), so d  f k +1 and d  f k +2 . Now f k +2 = f k + f k +1 , so f k = f k +2 f k +1 . Hence d  f k by Prop. 2.11.ii. Since d  f k and d  f k +1 , it follows from the Induction Hypothesis that d = 1, since gcd( f k , f k +1 ) = 1. (Alternatively, we could have said that gcd( f k +1 , f k +2 ) = gcd( f k +1 , f k +1 + f k ) = gcd( f k +1 , 1( f k +1 ) + f k ) = gcd( f k +1 , f k ) by Proposition 2.21, and then used the Induction Hypothesis.) Therefore, the result is true for n = k + 1....
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 Spring '08
 Vanderburgh

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