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Assignment 3 Solutions - MATH 135 Fall 2010 Solution of...

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MATH 135, Fall 2010 Solution of Assignment #3 Problem 1 . Prove that n i =1 (2 i - 1) 2 = n (2 n - 1)(2 n +1) 3 for all n 1. Solution. When n = 1 we have n i =1 (2 i - 1) 2 = 1 2 = 1 and n (2 n - 1)(2 n +1) 3 = 1 · 1 · 3 3 = 1, so the claim is true when n = 1. Let k 1 and suppose the claim is true when n = k , that is suppose that k i =1 (2 i - 1) 2 = k (2 k - 1)(2 k +1) 3 . Then when n = k + 1 we have = k +1 X i =1 (2 i - 1) 2 = k X i =1 (2 i - 1) 2 + (2 k + 1) 2 = k (2 k - 1)(2 k + 1) 3 + (2 k + 1) 2 = (2 k + 1) k (2 k - 1) 3 + (2 k + 1) = (2 k + 1) k (2 k - 1) + 3(2 k + 1) 3 = (2 k + 1)(2 k 2 - k + 6 k + 3) 3 = (2 k + 1)(2 k 2 + 5 k + 3) 3 = (2 k + 1)( k + 1)(2 k + 3) 3 Thus the claim is true when n = k + 1. By Mathematical Induction, the claim is true for all n 0. Problem 2 . Let a 0 = 1 and a 1 = 1 and for n 2 let a n = 2 a n - 1 + 2 a n - 2 . Show that we have a n = 1 2 ( (1 + 3) n + (1 - 3) n ) for all n 0. Solution. When n = 0 we have a n = a 0 = 1 and 1 2 ( (1 + 3) n + (1 - 3) n ) = 1 2 ( (1 + 3) 0 + (1 - 3) 0 ) = 1 2 ( 1 - 1 ) = 0, so the claim is true when n = 0. When n = 1 we have a n = a 1 = 1 and 1 2 ( (1+ 3) n +(1 - 3) n ) = 1 2 ( (1+ 2) 1 +(1 - 3) 1 ) = 1 2 ( 1+ 3+1 - 3 ) = 1, so the claim is true
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