Explanation Given information: Suppose that the sequences s 0 , s 1 , s 2 ,. .. and t 0 , t 1 , t 2 ,. .. both satisfy the same second-order linear homogeneous recurrence relation with constant coefficients: s k = 5 s k − 1 − 4 s k − 2 for all integers k ≥ 2 t k = 5 t k − 1 − 4 t k − 2 for all integers k ≥ 2 Proof: Let k be an integer such that k ≥ 2. 2 s k + 3 t k = 2 ⋅ ( 5 s k − 1 − 4 s k − 2 ) + 3 t k [ s k = 5 s k − 1 − 4 s k − 2 ] = 2 ⋅ ( 5 s k − 1 − 4 s k − 2 ) + 3 ⋅ ( 5 s k − 1 − 4 s k − 2 ) [ s k = 5 s k − 1 − 4 s k − 2 ] = 2 ⋅ 5 ⋅ s k − 1 − 2 ⋅ 4 ⋅ s k − 2 + 3 ⋅

5th Edition

ISBN: 9781337694193

Author: EPP, Susanna S.

Publisher: Cengage Learning,

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Chapter 5.8, Problem 18ES

To determine

To prove:

Show that 2sk+3tk=5(2sk−1+3tk−1)−4(2sk−2+3tk−2) for all integers k≥2.

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Chapter 5.8, Problem 17ES

Chapter 5.8, Problem 19ES

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Chapter 5 Solutions

Discrete Mathematics With Applications

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To prove: Show that 2 s k + 3 t k = 5 ( 2 s k − 1 + 3 t k − 1 ) − 4 ( 2 s k − 2 + 3 t k − 2 ) for all integers k ≥ 2 .

Solution Summary: The author explains how to prove that k is an integer such that it satisfies the same second-order linear homogeneous recurrence relation with constant coefficients.

Videos

To prove:

Show that 2sk+3tk=5(2sk−1+3tk−1)−4(2sk−2+3tk−2) for all integers k≥2.

Want to see the full answer?

Chapter 5 Solutions

To prove: Show that 2 s k + 3 t k = 5 ( 2 s k − 1 + 3 t k − 1 ) − 4 ( 2 s k − 2 + 3 t k − 2 ) for all integers k ≥ 2 .

Solution Summary: The author explains how to prove that k is an integer such that it satisfies the same second-order linear homogeneous recurrence relation with constant coefficients.

Videos

To prove: Show that 2sk+3tk=5(2sk−1+3tk−1)−4(2sk−2+3tk−2) for all integers k≥2.

Want to see the full answer?

Chapter 5 Solutions

To prove:

Show that 2sk+3tk=5(2sk−1+3tk−1)−4(2sk−2+3tk−2) for all integers k≥2.