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.

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Description: 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 forevery integer k 1.Let k be any integer with k2 1. Substitute k and k – 1 in pla

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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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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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.

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