Result Proof A sequence {an} is defined recursively by a₁ = 1, a₂ = 3 and an = 2an-1 Then an = 2n - 1 for all ne N. -an-2 for n ≥ 3. We proceed by induction. Since a₁ = 2 · 1 − 1 = 1, the formula holds for n = 1. Assume for an arbitrary positive integer k that a; = 2i - 1 for all integers i with 1 ≤ i ≤k. We show that ak+1 = 2(k+ 1) − 1 = 2k + 1. If k = 1, then ak+1 = a₂ = 2·1+1 = 3. Since a2 = 3, it follows that ak+1 = 2k + 1 when k = 1. Hence, we may assume that k≥ 2. Since k + 1 ≥ 3, it follows that ak+1 = 2ak - ak-1 = 2(2k – 1) – (2k − 3) = 2k + 1, which is the desired result. By the Strong Principle of Mathematical Induction, an 2n 1 for all n € N. =
Result Proof A sequence {an} is defined recursively by a₁ = 1, a₂ = 3 and an = 2an-1 Then an = 2n - 1 for all ne N. -an-2 for n ≥ 3. We proceed by induction. Since a₁ = 2 · 1 − 1 = 1, the formula holds for n = 1. Assume for an arbitrary positive integer k that a; = 2i - 1 for all integers i with 1 ≤ i ≤k. We show that ak+1 = 2(k+ 1) − 1 = 2k + 1. If k = 1, then ak+1 = a₂ = 2·1+1 = 3. Since a2 = 3, it follows that ak+1 = 2k + 1 when k = 1. Hence, we may assume that k≥ 2. Since k + 1 ≥ 3, it follows that ak+1 = 2ak - ak-1 = 2(2k – 1) – (2k − 3) = 2k + 1, which is the desired result. By the Strong Principle of Mathematical Induction, an 2n 1 for all n € N. =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please show how the highlighted part works step by step, I dont get the substitution, how do we get ak=2k-1 and
ak-1=2k-3?
Transcribed Image Text:Result
Proof
A sequence {an} is defined recursively by
a₁ = 1, a₂ = 3 and an = 2an-1
Then an = 2n - 1 for all ne N.
-an-2 for n ≥ 3.
We proceed by induction. Since a₁ = 2∙1 − 1 = 1, the formula holds for n = 1.
Assume for an arbitrary positive integer k that a; = 2i - 1 for all integers i with 1 ≤ i ≤k.
We show that ak+1 = 2(k+ 1) − 1 = 2k + 1. If k = 1, then ak+1 = a₂ = 2·1+1 = 3.
Since a2 = 3, it follows that ak+1 = 2k + 1 when k = 1. Hence, we may assume that
k≥ 2. Since k + 1 ≥ 3, it follows that
ak+1 = 2ak-ak-1 = 2(2k − 1) – (2k − 3) = 2k + 1,
=
which is the desired result. By the Strong Principle of Mathematical Induction, an
2n 1 for all n € N.
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