Homework 7 Solutions - MATH 55 2\/26 DISCUSSION QUESTIONS 1 Prove the following statements for all positive integers n by using induction(a 12 22 n2 =

Homework 7 Solutions - MATH 55 2/26 DISCUSSION QUESTIONS 1...

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MATH 55 2/26 DISCUSSION QUESTIONS 1. Prove the following statements for all positive integers n by using induction. (a) 1 2 + 2 2 + ... + n 2 = n ( n +1)(2 n +1) 6 (b) 6 divides n 3 - n (c) 1 2 - 2 2 + 3 2 - ... + ( - 1) n - 1 n 2 = ( - 1) n - 1 n ( n +1) 2 Solution: (a) I’ll leave the base case n = 1 to you. Suppose that P ( k ) is true. Then 1 2 + 2 2 + ... + k 2 + ( k + 1) 2 = k ( k + 1)(2 k + 1) 6 + ( k + 1) 2 = ( k + 1) k (2 k + 1) + 6 k + 6 6 = ( k + 1) 2 k 2 + 7 k + 6 6 = ( k + 1)( k + 2)(2 k + 3) 6 as we wanted to show. (b) I’ll leave the base case n = 1 to you (notice that 6 divides 0). For the induction step, assume P ( k ), so k 3 - k = 6 m for some integer m . Then ( k + 1) 3 - ( k + 1) = k 3 + 3 k 2 + 3 k + 1 - k - 1 = ( k 3 - k ) + 3 k 2 + 3 k = 6 m = (3 k 2 + 3 k ) , so we need to show that 6 divides 3 k 2 + 3 k . Well, k 2 + k is even since k and k 2 have the same parity (we showed this a long time ago), so k 2 + k = 2 y for some integer y . Then 3 k 2 + 3 k = 6 y so it’s a multiple of 6 as needed. (c) I’ll leave the base case n = 1 to you again. Assume P ( k ). Then 1 2 - 2 2 + 3 2 - ... + ( - 1) k - 1 k 2 + ( - 1) k ( k + 1) 2 = ( - 1) k - 1 k ( k + 1) 2 + ( - 1) k ( k + 1) 2 = ( - 1) k ( k + 1) 2 - k ( k + 1) 2 = ( - 1) k 2 k 2 + 4 k + 2 - k 2 - k 2 = ( - 1) k k 2 + 3 k + 2 2 = ( - 1) k ( k + 1)( k + 2) 2 which is what we wanted to show.
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