Explanation Consider the homogeneous recurrence relation a n = 5 a n − 1 − 2 a n − 2 with a n = r a n − 1 + s a n − 2 to get r = 5 and s = − 2 . The characteristic polynomial is given by x 2 − 5 x + 2 . The roots of the above polynomial can be calculated as x 2 − 5 x + 2 = 0 x = 5 ± 25 − 4 × 2 2 x = 5 ± 17 2 Thus, a n = c 1 ( 5 + 17 2 ) n + c 2 ( 5 − 17 2 ) n . p n is any particular solution of the recurrence relation a n = r a n − 1 + s a n − 2 + f ( n ) and q n is the general solution of the homogeneous recurrence relation a n = r a n − 1 + s a n − 2 . Then, a n = p n + q n is the general solution of the recurrence relation a n = r a n − 1 + s a n − 2 + f ( n ) . The initial conditions determine the constants in q n . For a n = 5 a n − 1 − 2 a n − 2 + 3 n 2 As f ( n ) = 3 n 2 , let us assume that p n = a + b n + c n 2 . As p n is the particular solution of the recurrence relation, it must satisfy the relation. So, p n = 5 a n − 1 − 2 a n − 2 + 3 n 2 . Substituting, p n = a + b n + c n 2 in the above equation. a + b n + c n 2 = 5 ( a + b ( n − 1 ) + c ( n − 1 ) 2 ) − 2 ( a + b ( n − 2 ) + c ( n − 2 ) 2 ) + 3 n 2 a + b n + c n 2 = 5 a + 5 b ( n − 1 ) + 5 c ( n − 1 ) 2 − 2 a − 2 b ( n − 2 ) − 2 c ( n − 2 ) 2 + 3 n 2 a + b n + c n 2 = 5 a + 5 b n − 5 b + 5 c ( n 2 + 1 − 2 n ) − 2 a − 2 b n + 4 b − 2 c ( n 2 + 4 − 4 n ) + 3 n 2 a + b n + c n 2 = 2 a + 3 b n − b + 3 n 2 + 5 c n 2 + 5 c − 10 n c − 2 c n 2 − 8 c + 8 n a + b n + c n 2 = 2 a − b − 3 c + 3 b n − 10 n c + 8 n + 3 n 2 + 3 c n 2 a + b n + c n 2 = ( 2 a − b − 3 c ) + ( 3 b − 10 c + 8 ) n + ( 3 + 3 c ) n 2 We get a = − 3 , b = − 3 2 , and c = − 3 2

3rd Edition

ISBN: 9780131679955

Author: Edgar G. Goodaire

Publisher: Prentice Hall

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Chapter 5.3, Problem 17E

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The solution of the recurrence relation an=5an−1−2an−2+3n2, n≥2, given a0=0, a1=3.

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Chapter 5.3, Problem 16E

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The solution of the recurrence relation a n = 5 a n − 1 − 2 a n − 2 + 3 n 2 , n ≥ 2 , given a 0 = 0 , a 1 = 3 .

Solution Summary: The author explains the solution of the homogeneous recurrence relation a_n=5

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The solution of the recurrence relation an=5an−1−2an−2+3n2, n≥2, given a0=0, a1=3.

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