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Unformatted text preview: MATH 239 Quiz — Winter 2009 SOLUTIONS No calculators or other aids may be used. Total marks: 20 1. (a) Let m and n be positive integers. Find an expression (you may use summation notation) for [ x m ](1 + 2 x ) n (1 + x 2 ) 2 . [5 marks] Solution [ x m ](1 + 2 x ) n (1 + x 2 ) 2 = [ x m ] n X j =0 n j 2 j x j ∞ X k =0 ( k + 1)( 1) k x 2 k ! = [ x m ] n X j =0 ∞ X k =0 ( 1) k ( k + 1) n j 2 j x j +2 k . Since we get an x m whenever j + 2 k = m , we may sum over all possible k and set j = m 2 k to get [ x m ] n X j =0 ∞ X k =0 ( 1) k ( k + 1) n j 2 j x j +2 k = b m/ 2 c X k =0 ( 1) k ( k + 1) n m 2 k 2 m 2 k . (b) Let S be the set of all subsets of the positive integers { 1 , 2 , 3 , . . . } . Define the weight function ω as follows. For the empty set {} , we let ω ( {} ) = 0 and for any nonempty set σ we let ω ( σ ) equal the largest element of the subset σ . For example, if σ = { 1 , 2 , 4 } , then ω ( σ ) = 4, since 4 is the largest element in...
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This note was uploaded on 08/09/2009 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Combinatorics, Integers

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