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quiz-2-solutions

quiz-2-solutions - MATH 239 Quiz Spring 2010 SOLUTIONS No...

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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 k =0 r k ( - x ) k j 0 5 + j - 1 j ( - 3 x 2 ) j = [ x n ] r k =0 j 0 r k 4 + j 4 ( - 1) k + j 3 j x 2 j + k = [ x n ] j 0 r 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 n/ 2 . 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 n/ 2 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: 0 k r, k n, n - k even r k 4 + ( n - k ) / 2 4 ( - 1) ( n + k ) / 2 3 ( n - k ) / 2 . 2. At a garage sale there are two cellphones costing \$5 and \$10 respectively, one MP3 player for \$15 and seven identical blank cd’s for \$1 each. Let

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