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Unformatted text preview: MATH 135 Assignment 6 This assignment is due at 8:30am on Wednesday March 3, in the drop boxes opposite the Tutorial Centre, MC 4067. 1. [8 marks] Let the sequence of integers c n be defined for n ≥ 0 as follows: c = 0, c 1 = 2, c 2 = 13, and c n = 4 c n 1 + 3 c n 2 + 18 c n 3 for all n ≥ 3. Prove that c n = 2 n + ( n 1)( 3) n for all n ≥ 0. Solution: We use induction on n . Base cases: For n = 0 we have c = 0 by definition, while 2 + (0 1)( 3) = 1 1 = 0. For n = 1 we have c 1 = 2 by definition, while 2 1 + (1 1)( 3) 1 = 2 + 0 = 2. Finally for n = 2 we have c 2 = 13 by definition, while 2 2 + (2 1)( 3) 2 = 4 + 9 = 13. Induction Hypothesis: Let k ≥ 2, and assume that c j = 2 j + ( j 1)( 3) j for all 0 ≤ j ≤ k . Induction step: Consider n = k + 1. Since n ≥ 3 we have by definition c n = c k +1 = 4 c k + 3 c k 1 + 18 c k 2 . Using the Induction Hypothesis applied to c k , c k 1 and c k 2 , we get c n = c k +1 = 4(2 k + ( k 1)( 3) k ) + 3(2 k 1 + ( k 2)( 3) k 1 ) + 18(2 k 2 + ( k 3)( 3) k 2 ) . Simplfying gives c k +1 = 2 k 2 ( 4(4) + 3(2) + 18) + ( 3) k 2 ( 4(9)( k 1) + 3( 3)( k 2) + 18( k 3) ....
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 Spring '08
 ANDREWCHILDS
 Binomial Theorem, Integers, Mathematical Induction, Natural number, Prime number, NK

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