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Result A sequence {an} is defined recursively by a₁ = 1, a₂ = 4 and an = Then an = n² for all n € N. 2an-1 -an-2 +2 for n ≥ 3.

Result A sequence {an} is defined recursively by a₁ = 1, a₂ = 4 and an = Then an = n² for all n € N. 2an-1 -an-2 +2 for n ≥ 3.

Advanced Engineering Mathematics
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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please show the following proof in more detail, I'm especially lost on the highlighted part, thank you in advance.

 

Result Proof A sequence {an) is defined recursively by a₁ = 1, a₂ = 4 and an = 2an-1 — an-2 +2 for n ≥ 3. Then an = n² for all n € N. We proceed by induction. Since a₁ = 1 = : 12, the formula holds for n = 1. Assume for an arbitrary positive integer k that ai i² for every integer i with 1 ≤ i ≤k. We show = that ak+1 = (k+ 1)². Since a2 = 4, it follows that ak+1 = (k+ 1)² when k = 1. Thus, we may assume that k ≥ 2. Hence, k + 1 ≥ 3 and so ak+1 = 2ak - ak-1 + 2 = 2k² − (k − 1)² + 2 = = 2k² − (k² − 2k + 1) + 2 = k² + 2k + 1 = (k + 1)². By the Strong Principle of Mathematical Induction, an = n² for all n € N.
Transcribed Image Text:Result Proof A sequence {an) is defined recursively by a₁ = 1, a₂ = 4 and an = 2an-1 — an-2 +2 for n ≥ 3. Then an = n² for all n € N. We proceed by induction. Since a₁ = 1 = : 12, the formula holds for n = 1. Assume for an arbitrary positive integer k that ai i² for every integer i with 1 ≤ i ≤k. We show = that ak+1 = (k+ 1)². Since a2 = 4, it follows that ak+1 = (k+ 1)² when k = 1. Thus, we may assume that k ≥ 2. Hence, k + 1 ≥ 3 and so ak+1 = 2ak - ak-1 + 2 = 2k² − (k − 1)² + 2 = = 2k² − (k² − 2k + 1) + 2 = k² + 2k + 1 = (k + 1)². By the Strong Principle of Mathematical Induction, an = n² for all n € N.
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