Hw1solutions - 2 Prove that 1 √ 1 1 √ 2 ·· 1 √ n ≥ √ n for all n ∈ N We again argue using induction The n = 1 base case follows

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: January 20, 2010 Assignment 1 1. Prove that n 3 + ( n + 1) 3 + ( n + 2) 3 is divisible by 9 for all n N . We prove this via induction. The base case of n = 1 is trivial to check since the given quantity is 36. To proceed via induction, we assume that the n - 1 case holds. That is, we assume that ( n - 1) 3 + n 3 + ( n + 1) 3 is divisible by 9. We need to use this to prove that n 3 + ( n + 1) 3 + ( n + 2) 3 is divisible by 9. To do so, note that ( n + 2) 3 = (( n - 1) + 3) 3 = ( n - 1) 3 + 9( n - 1) 2 + 27( n - 1) + 27 . Then, n 3 + ( n + 1) 3 + ( n + 2) 3 = [ n 3 + ( n + 1) 3 + ( n - 1) 3 ] + [9( n - 1) 2 + 27( n - 1) + 27] . The divisibility by 9 of the quantity in the first set of brackets follows from the inductive hypothesis, while the divisibility by 9 of the second quantity follows by inspecting the coefficients.
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Unformatted text preview: 2. Prove that 1 / √ 1 + 1 / √ 2 + ··· + 1 / √ n ≥ √ n for all n ∈ N . We, again, argue using induction. The n = 1 base case follows trivially as both sides are equal to 1 in that case. We then assume that the result holds for n replaced by k-1 and seek to use that to prove the n = k case. Here, by this inductive hypothesis, we have 1 / √ 1 + 1 / √ 2 + ··· + 1 / √ k-1 + 1 / √ k ≥ √ k-1 + 1 √ k . Thus, it will suffice to show that √ k-1 + 1 √ k ≥ √ k. Note that p k ( k-1) ≥ p ( k-1) 2 = k-1 , for k ≥ 2 . Thus, p k ( k-1) + 1 ≥ k in this range. Or √ k-1 + 1 √ k ≥ √ k, which is the desired inequality. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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