quiz-1-and-4-solutions

# quiz-1-and-4-solutions - Math 239 Quiz Solutions 1(5 points...

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Math 239: Quiz Solutions 1. (5 points) Assume r and n are non-negative integers. Express the coefﬁcient of x n in the following series as a summation: (1 - 3 x ) r (1 + 5 x 2 ) - 10 . Solution: Using the binomial theorem we have (1 - 3 x ) r = X i 0 ( - 3) i ˆ r i ! x i and (1 + 5 x 2 ) - 10 = X j 0 5 j ˆ - 10 j ! x 2 j . Hence setting i = n - 2 j and using the product rule, we get [ x n ](1 - 3 x ) r (1 + 5 x 2 ) - 10 = X 0 j n /2 ( - 3) n - 2 j 5 j ˆ r n - 2 j - 10 j ! = ( - 3) n X 0 j n /2 ± 5 9 j ˆ r n - 2 j - 10 j ! 2. (5 points) At a bake sale, there is one cake for \$12, two pies costing \$6 and \$8, and 10 mufﬁns at \$1 each. Let b n be the number of ways that you can spend \$n. Find the gen- erating function n 0 b n x n in the form of a rational function. Give full justiﬁcation for your answer. Solution: Deﬁne the four sets S 1 = {0,12}, S 2 = {0,6}, S 3 = {0,8}, S 4 = {0,1,. ..,10}. Then our spending decisions correspond to the elements of

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quiz-1-and-4-solutions - Math 239 Quiz Solutions 1(5 points...

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