# 12 - The University of Sydney MATH 1004 Second Semester Discrete Mathematics 2011 Tutorial 12 Week 13 1(i Show that xn = 6 2n 4 is a solution to

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The University of Sydney MATH 1004 Second Semester Discrete Mathematics 2011 Tutorial 12 Week 13 1. ( i ) Show that x n = 6 · 2 n - 4 is a solution to the recurrence relation x n = 3 x n - 1 - 2 x n - 2 . ( ii ) Show that x n = (3 n +1 - 1) / 2 is a solution to the recurrence relation x n +1 - x n = 3 n +1 . ( iii ) Show that x n = n ! is a solution to the recurrence relation x n - n ( n - 1) x n - 2 = 0 . 2. Solve the following recurrence relations: ( i ) x n - 5 x n - 1 + 6 x n - 2 = 0 , n 2 , x 0 = 3 , x 1 = 7 ( ii ) x n - 8 x n - 1 + 16 x n - 2 = 0 , n 2 , x 0 = 3 , x 1 = 20 . 3. Solve the following recurrence relations: ( i ) x n = 10 x n - 1 - 25 x n - 2 , for n 2, where x 0 = - 1 and x 1 = 5. ( ii ) x n +3 - 6 x n +2 + 11 x n +1 - 6 x n = 0, for n 0, where x 0 = 1, x 1 = 0 and x 2 = - 1. 4. Solve the following recurrence relations: ( i ) x n = 4 x n - 1 - 3 x n - 2 , where x 0 = 1 and x 1 = 2. ( ii ) x n = 3 x n - 1 - 3 x n - 2 + x n - 3 , where x 0 = 0, x 1 = 1 and x 2 = 3.

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The University of Sydney MATH 1004 Second Semester Discrete Mathematics 2011 Problem Set 12 1. Solve the following recurrence relations: ( i ) x n + 6 x n - 1 - 7 x n - 2 = 0 for all n 2, with x 0 = 1 and
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## This note was uploaded on 09/01/2011 for the course YEAR 1 taught by Professor Various during the Three '11 term at University of Sydney.

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12 - The University of Sydney MATH 1004 Second Semester Discrete Mathematics 2011 Tutorial 12 Week 13 1(i Show that xn = 6 2n 4 is a solution to

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