Solutions_2_W - MAT 312/AMS 351, Summer 2011 Solutions to...

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MAT 312/AMS 351, Summer 2011 Solutions to Homework Assignment 2 Maximal grade for HW2: 100 points Section 1.2. 2. (15 points) Prove that for all positive integers n , 1 + 2 2 + 3 2 + ... + n 2 = n ( n + 1)(2 n + 1) 6 . (1) Solution: We prove this statement using induction by n . Base. We have to check the equation for n = 1: 1(1 + 1)(2 · 1 + 1) 6 = 2 · 3 6 = 1 . Step. Suppose that we proved the formula (1) for some number n , let us prove it for n + 1. We have: 1+2 2 +3 2 + ... + n 2 +( n +1) 2 = (1+2 2 +3 2 + ... + n 2 )+( n +1) 2 = n ( n + 1)(2 n + 1) 6 +( n +1) 2 = n + 1 6 ( n (2 n +1)+6( n +1)) = n + 1 6 (2 n 2 +7 n +6) = n + 1 6 ( n +2)(2 n +3) = ( n + 1)(( n + 1) + 1)(2( n + 1) + 1) 6 . The last formula is precisely the right hand side of (1) with n replaced by ( n + 1), so the equation (1) is true for ( n + 1) as well. 3. (15 points) The Fibonacci sequence is the sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 ... where each term is the sum of the two preceding terms. Show that every two successive terms of the Fibonacci sequence are relatively prime. 1
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Solution: We can define the Fibonacci sequence as follows: F 1 = F 2 = 1 , F n = F n - 1 + F n - 2 for n > 2 . Let us prove that
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Solutions_2_W - MAT 312/AMS 351, Summer 2011 Solutions to...

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