園 •5) Given the sequence defined by the following Bo=12 recurrence relation: . By 29 nonnegative Integer n. •B₁ = s. bi-1-6 · bi-z for izz Prove that bn = 5 37.2" for any
園 •5) Given the sequence defined by the following Bo=12 recurrence relation: . By 29 nonnegative Integer n. •B₁ = s. bi-1-6 · bi-z for izz Prove that bn = 5 37.2" for any
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.2: Trigonometric Equations
Problem 75E
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Transcribed Image Text:Complete the basis step of the proof.
What is the inductive hypothesis?
What do you need to show in the inductive step of the proof?
Complete the inductive step of the proof.
Transcribed Image Text:•5) Given the
∙Bo=12
·B₂ = 29
1
sequence defined by the following
• Bi = S · bi - 1-6-be-2 for izz
19
Prove
that bn = S⋅ 3" +7=2" for any
recurrence
relation:
nonnegative
Integer n.
12
h = o
bo
5.3° +7-(-2)° = S-3
n = 1
• 1 + 7.1 = S + 7 = 12
n-2
b₁ = 29
5·3' + 7· (-2)' = 5.3+7+2 = 15+14= 29
* = 5.3* +7.(-2)"
x=5.31-1
• bk
+ 7.2K-
K-1)
61+1 = 5 + ( 5.3*+7+ (-2) *) -6. (5.3+- +70 (-2)*-
bk+2= 25.3k +35. (-2) * -30.3k-1-42-(-2) K-
1-2
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