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Result For every integer n ≥ 5, 2">n². Proof We proceed by induction. Since 25 > 5², the inequality holds for n = 5. Assume that 2k > k² where k ≥ 5. We show that 2k+1 > (k+ 1)². Observe that 2k+1 = 2.2k > 2k² = k² + k² ≥ k² + 5k = k² + 2k + 3k ≥ k² + 2k + 15 > k² + 2k + 1 = (k+1)². Therefore, 2k+1> (k + 1)². By the Principle of Mathematical Induction, 2" >n² for every integer n ≥ 5.

Result For every integer n ≥ 5, 2">n². Proof We proceed by induction. Since 25 > 5², the inequality holds for n = 5. Assume that 2k > k² where k ≥ 5. We show that 2k+1 > (k+ 1)². Observe that 2k+1 = 2.2k > 2k² = k² + k² ≥ k² + 5k = k² + 2k + 3k ≥ k² + 2k + 15 > k² + 2k + 1 = (k+1)². Therefore, 2k+1> (k + 1)². By the Principle of Mathematical Induction, 2" >n² for every integer n ≥ 5.

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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I understand the proof until we get to the highlighted part, please explain it in more steps, thank you in advance.

 

Result For every integer n ≥ 5, 2">n². Proof We proceed by induction. Since 25 > 5², the inequality holds for n = 5. Assume that 2kk² where k≥ 5. We show that 2k+¹ > (k+ 1)². Observe that 2k+1 = 2.2k > 2k² = k² +k² > k² + 5k = k² + 2k + 3k ≥k² + 2k +15 - k² + 2k + 1 = (k + 1)². Therefore, 2k+¹ > (k+ 1)². By the Principle of Mathematical Induction, 2″ > n² for every integer n ≥ 5.
Transcribed Image Text:Result For every integer n ≥ 5, 2">n². Proof We proceed by induction. Since 25 > 5², the inequality holds for n = 5. Assume that 2kk² where k≥ 5. We show that 2k+¹ > (k+ 1)². Observe that 2k+1 = 2.2k > 2k² = k² +k² > k² + 5k = k² + 2k + 3k ≥k² + 2k +15 - k² + 2k + 1 = (k + 1)². Therefore, 2k+¹ > (k+ 1)². By the Principle of Mathematical Induction, 2″ > n² for every integer n ≥ 5.
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