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Unformatted text preview: 1 MATH239 MIDTERM EXAMINATION July 2 2009, 4:306:30pm INSTRUCTIONS 1. Fill in the following and check your section: Name: I.D.#: s Section 1 (M. Pei) s Section 2 (D. Jao) s Section 3 (P. Haxell) 2. No aids are allowed. 3. A complete paper has 9 pages, including this cover page. Check that you have them all. 4. You may assume the result of any theorem proved in the lectures or on assignments. State clearly which results you are using. 6. Show all your work. If the space is insu±cient, use the back of the page and indicate where your solution continues. For Marker Only 1 / 9 2 /10 3 / 8 4 /12 5 / 8 6 / 8 7 / 5 Σ /60 2 1. (a) [4 marks] Determine the given coefcient. (You do not need to evaluate binomial coefcients or exponentials explicitly.) [ x 9 ] p 5 x 2 (12 x ) 4 + (1 + 3 x ) 11 P . Solution: We Frst write [ x 9 ] p 5 x 2 (12 x ) 4 + (1 + 3 x ) 11 P = [ x 9 ] 5 x 2 (12 x ) 4 + [ x 9 ](1 + 3 x ) 11 and evaluate each summand separately. Using the rules ±or manipulating coefcients o± power series, together with the negative exponent binomial theorem, we have [ x 9 ] 5 x 2 (12 x ) 4 = 5[ x 7 ] 1 (12 x ) 4 = 5[ x 7 ] s k ≥ p 4 + k1 k P (2 x ) k = 5 · 2 7 · p 4 + 71 7 P = 5 · 2 7 · p 10 7 P . ²or the second part, we use the binomial theorem and obtain [ x 9 ](1 + 3 x ) 11 = [ x 9 ] s k ≥ p 11 k P (3 x ) k = 3 9 p 11 9 P Hence [ x 9 ] p 5 x 2 (12 x ) 4 + (1 + 3 x ) 11 P = 5 · 2 7 · p 10 7 P + 3 9 · p 11 9 P = 1159365 . (Note: Numerical evaluation is not required ±or ±ull credit.) 3 (b) [5 marks] Let k be a positive integer. Prove that p 2 k k P = k s j =0 p k j P 2 . Solution: Using the binomial theorem and the formula for the coeFcients of the product of two power series, we ±nd that p 2 k k P = [ x k ](1 + x ) 2 k = [ x k ](1 + x ) k (1 + x ) k = k s j =0 ([ x j ](1 + x ) k )([ x kj ](1 + x ) k ) = k s j =0 p k j Pp k kj P = k s j =0 p k j P 2 , where the last line follows from the identity ( n k ) = ( n nk ) . Alternate solution: The left hand side represents the number of ksubsets of a 2 kset. Partition the 2 kset into any partition A ∪ B where A and B each have k elements....
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 Spring '09
 M.PEI
 Math, Combinatorics, Recurrence relation, Fibonacci number, Generating function, D. Jao, P. Haxell

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