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.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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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