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Let bo, b,, b,, ... be defined by the formula b, = 4" for every integer n 2 0. Fill in the blanks to show that bo, b,, b2, ... satisfies the recurrence relation b, 4b,1 for every integer k 2 1. Let k be any integer with k > 1. Substitute k and k - 1 in place of n, and apply the definition of bo, b,, b,, ... to both b, and b,1. The result is b, = 4* (*) and b-1= (**) for every integer k 2 1. It follows that for every integer k > 1, 4bk - 1 = by substitution from ? by basic algebra by substitution from ? v Thus, bo, b,, b2,... satisfies the given recurrence relation.

Let bo, b,, b,, ... be defined by the formula b, = 4" for every integer n 2 0. Fill in the blanks to show that bo, b,, b2, ... satisfies the recurrence relation b, 4b,1 for every integer k 2 1. Let k be any integer with k > 1. Substitute k and k - 1 in place of n, and apply the definition of bo, b,, b,, ... to both b, and b,1. The result is b, = 4* (*) and b-1= (**) for every integer k 2 1. It follows that for every integer k > 1, 4bk - 1 = by substitution from ? by basic algebra by substitution from ? v Thus, bo, b,, b2,... satisfies the given recurrence relation.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 57EQ
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Let bo, b,, b,, ... be defined by the formula b, = 4" for every integer n 2 0. Fill in the blanks to show that bo, b,, bn, ... satisfies the recurrence relation b, 4b, for every integer k 21. Let k be any integer with k > 1. Substitute k and k - 1 in place of n, and apply the definition of bo, b,, b,, ... to both b, and b,1. The result is b = 4 (*) and bk-1 (**) for every integer k 2 1. It follows that for every integer k > 1, 4bk -1 = by substitution from ? v by basic algebra by substitution from ? v Thus, bo, b,, b2,... satisfies the given recurrence relation.
Transcribed Image Text:Let bo, b,, b,, ... be defined by the formula b, = 4" for every integer n 2 0. Fill in the blanks to show that bo, b,, bn, ... satisfies the recurrence relation b, 4b, for every integer k 21. Let k be any integer with k > 1. Substitute k and k - 1 in place of n, and apply the definition of bo, b,, b,, ... to both b, and b,1. The result is b = 4 (*) and bk-1 (**) for every integer k 2 1. It follows that for every integer k > 1, 4bk -1 = by substitution from ? v by basic algebra by substitution from ? v Thus, bo, b,, b2,... satisfies the given recurrence relation.
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