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Unformatted text preview: MATH 239 Quiz — Spring 2010 — SOLUTIONS No calculators or other aids may be used. Each of the 4 questions is worth 5 marks. 1. Determine the following coefficient, as a summation, where r and n are nonnegative integers. [ x n ](1 x ) r (1 + 3 x 2 ) 5 Solution [ x n ](1 x ) r (1 + 3 x 2 ) 5 = [ x n ] r X k =0 r k ( x ) k ! X j ≥ 5 + j 1 j ( 3 x 2 ) j ! = [ x n ] r X k =0 X j ≥ r k 4 + j 4 ( 1) k + j 3 j x 2 j + k = [ x n ] X j ≥ r X k =0 r k 4 + j 4 ( 1) k + j 3 j x 2 j + k . To get x n we need 2 j + k = n , or k = n 2 j . Then k ≥ 0 is equivalent to 2 j ≤ n , or j ≤ b n/ 2 c . If k > r , the binomial coefficient ( r k ) is 0 in any case so the condition k ≤ r may be ignored. So the answer is b n/ 2 c X j =0 r n 2 j 4 + j 4 ( 1) n j 3 j . Note there are several alternative solutions. For instance we could sum over k , using j = ( n k ) / 2. But then only k congruent to n (mod 2) should be included in the summation and of course the expression is different: X ≤ k ≤ r, k ≤ n, n k even...
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This note was uploaded on 11/26/2011 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Math, Integers

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