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Unformatted text preview: SOLUTIONS FOR HOMEWORK 5 2.18. This is a bonus problem . Write our polynomial as p ( x ) = ∑ n k =0 a k x k . Denote the set of even and odd integers between 0 and n by E and O , respectively. Then p ( x ) = ∑ k ∈ E a k x k + ∑ k ∈ E a k x k . Let A = ∑ k ∈ E a k and B = ∑ k ∈ O a k . Then p (1) = A + B , while p ( 1) = A B . Therefore, p (1) p ( 1) = ( A + B )( A B ) = A 2 B 2 . 3.15. We have to show that ∑ n i =1 ( 1) i i 2 = ( 1) n n ( n + 1) / 2. The basic step consists of verifying the above equality for n = 1. Then we get ( 1) 1 · 1 2 = ( 1) 2 · 1 · (1 +1) / 2, which is true. To perform the inductive step, we show that, for any n ∈ N , ∑ n i =1 ( 1) i i 2 = ( 1) n n ( n +1) / 2 implies ∑ n +1 i =1 ( 1) i i 2 = ( 1) n ( n + 1)( n + 2) / 2 Indeed, by the induction hypothesis, n +1 summationdisplay i =1 ( 1) i i 2 = n summationdisplay i =1 ( 1) i i 2 + ( 1) n +1 ( n + 1) 2 = ( 1) n n ( n + 1) 2 + ( 1) n +1 ( n + 1) 2 = ( 1) n +1 ( n + 1) parenleftBig n 2 + ( n + 1) parenrightBig = ( 1) n +1 ( n + 1) n + 2 2 , which completes the proof. 3.22. Our goal is to establish that statement P ( n ) = ( ∀ a 1 ,... ,a n ∈ R )(  n summationdisplay i =1 a i  ≥ n summationdisplay i =1  a i  ) is true for every n . The basic step is easy: P (1) is true. The inductive step consists of showing that if, for any a 1 ,... ,a n ∈ R we have  ∑ n i =1 a i  ≤ ∑ n i =1  a i  , then also  ∑ n +1 i =1 a i  ≤ ∑ n +1 i =1  a i  (for any a 1 ,... ,a n +1 ∈ R . Let a = ∑ n i =1 a i . Recall Triangle Inequality (Proposition 1.3): for any x,y ∈ R ,  x + y  ≤  x  +  y  . Then  n +1 summationdisplay i =1 a i  =  n summationdisplay i =1 a i + a n +1  =  a + a n +1  ....
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This note was uploaded on 01/27/2010 for the course MATH 347 taught by Professor ? during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 ?
 Math, Integers

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