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Let bo, b,, b2,... be defined by the formula b, = 4" for every integern2 0. Fill in the blanks to show that bo, b,, b,, . satisfies the recurrence relation b = 4bk - 1 for every integer k 1. Let k be any integer with k2 1. Substitute k and k - 1 in place of n, and apply the definition of b, b,, b,, .. to both b, and bK The result is 1 b = 4k (*) and bk -1= (**) for every integer k 2 1. It follows that for every integer k 2 1, 4bk - 1 by substitution from ? v by basic algebra by substitution from ? Thus, bo, b,, b2! satisfies the given recurrence relation.

Let bo, b,, b2,... be defined by the formula b, = 4" for every integern2 0. Fill in the blanks to show that bo, b,, b,, . satisfies the recurrence relation b = 4bk - 1 for every integer k 1. Let k be any integer with k2 1. Substitute k and k - 1 in place of n, and apply the definition of b, b,, b,, .. to both b, and bK The result is 1 b = 4k (*) and bk -1= (**) for every integer k 2 1. It follows that for every integer k 2 1, 4bk - 1 by substitution from ? v by basic algebra by substitution from ? Thus, bo, b,, b2! satisfies the given recurrence relation.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let bo, b,, b2... be defined by the formula b, = 4" for every integern2 0. Fill in the blanks to show that bo, b,, b,, . satisfies the recurrence relation b = 4bk - 1 for every integer k 1. Let k be any integer with k2 1. Substitute k and k – 1 in place of n, and apply the definition of b, b,, b,, . to both b̟ and b, The result is 1 b, = 4k (*) and bk -1 = (**) for every integer k 2 1. It follows that for every integer k 2 1, 4bk -1 = by substitution from ? v by basic algebra by substitution from ? Thus, bo, b,, b2! ... satisfies the given recurrence relation.
Transcribed Image Text:Let bo, b,, b2... be defined by the formula b, = 4" for every integern2 0. Fill in the blanks to show that bo, b,, b,, . satisfies the recurrence relation b = 4bk - 1 for every integer k 1. Let k be any integer with k2 1. Substitute k and k – 1 in place of n, and apply the definition of b, b,, b,, . to both b̟ and b, The result is 1 b, = 4k (*) and bk -1 = (**) for every integer k 2 1. It follows that for every integer k 2 1, 4bk -1 = by substitution from ? v by basic algebra by substitution from ? Thus, bo, b,, b2! ... satisfies the given recurrence relation.
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