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Unformatted text preview: SOLUTIONS FOR MIDTERM 1 (10 points) : Prove that, for any positive integer n , n +1 summationdisplay k =1 k 3 = ( n + 1) 2 ( n + 2) 2 4 . Define Proposition P m : ∑ m k =1 k 3 = m 2 ( m + 1) 2 / 4, and show that P m holds for any m ∈ N (the equality in the statement of the problem is P n +1 ). We use induction. The base is easy to establish for m = 1. Indeed, P 1 states that 1 3 = 1 2 (1 + 1) 2 / 4, which is true. The induction step consists of proving that P n implies P n +1 . To this end, assume ∑ n k =1 k 3 = n 2 ( n + 1) 2 / 4, and show that ∑ n +1 k =1 k 3 = ( n + 1) 2 ( n + 2) 2 / 4. We have: n +1 summationdisplay k =1 k 3 = n summationdisplay k =1 k 3 + ( n + 1) 3 = n 2 ( n + 1) 2 4 + ( n + 1) 3 = ( n + 1) 2 4 parenleftBig n 2 + 4( n + 1) parenrightBig = ( n + 1) 2 4 parenleftBig n 2 + 4 n + 4 parenrightBig = ( n + 1) 2 ( n + 2) 2 4 . 2 (10 points) : Suppose S is a subset of (0 , ∞ ), bounded above. Let S 1 = { s 1  s ∈ S } . Prove that S 1 is bounded below, and inf...
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This note was uploaded on 05/08/2010 for the course MATH 140a taught by Professor Staff during the Winter '08 term at UC Irvine.
 Winter '08
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