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assignment7_sol - MATH 135 Assignment 7 Solutions Hand-In...

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MATH 135 Fall 2005 Assignment 7 Solutions Hand-In Problems Exercise 4-28: 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 0 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 4-30: 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.

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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 4-60: 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. Therefore, by the Principle of Mathematical Induction, gcd( f n , f n +1 ) = 1 for all n P . Problem 4-67: Prove that f n +5 3 f n (mod 5) for all n P Solution 1: Using the recurrence relation for the Fibonacci sequence repeatedly, f n +5 = f n +4 + f n +3 = ( f n +3 + f n +2 ) + f n +3 = 2 f n +3 + f n +2 = (2 f n +2 + 2 f n +1 ) + f n +2 = 3 f n +2 + 2 f n +1 = (3 f n +1 + 3 f n ) + 2 f n +1 = 5 f n +1 + 3 f n 5 f n +1 + 3 f n (mod 5) 3 f n (mod 5) Solution 2 We shall prove the result by strong induction on n .
Base Cases If n = 1, then f n +5 = f 6 = 8 and f n = f 1 = 1, and 8 3(1) (mod 5), so f n +5 3 f n (mod 5).

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assignment7_sol - MATH 135 Assignment 7 Solutions Hand-In...

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