ult Proof A sequence {an} is defined recursively by a₁ = 1, a₂ = 3 and an = 2an-1 − An-2 for n ≥ 3. Then an = 2n - 1 for all n € N. 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 tha 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.

ult Proof A sequence {an} is defined recursively by a₁ = 1, a₂ = 3 and an = 2an-1 − An-2 for n ≥ 3. Then an = 2n - 1 for all n € N. 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 tha 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

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Question

Please explain the following proof by the strong induction principle, show why each step is taken, I understand induction I just really dont understand what we're trying to do here, so please explain in detail. thank you in advance.

Result
A sequence {an} is defined recursively by
a1 = = 1, a₂ =
= 3 and an
Then an = 2n 1 for all n € N.
=
2an-1
an-2 for n ≥ 3.
Proof 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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