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60 Use Induction to prove that 10n < 3" for all integers n > 4. Proof: The Basis Step: n = 4 10n 40 = (integer) 3n 81 (integer) Thus, the basis step The Induction Step: holds (holds/fails) Assume that 10k <3 is true for some integer k > 4 Then 3k+1 = 3. 3^k 20k 3. 10k > 10k+ where the last inequality is true because k > 4 20k ≥ 80. = 10k+ = 10(k + 1), which means that

60 Use Induction to prove that 10n < 3" for all integers n > 4. Proof: The Basis Step: n = 4 10n 40 = (integer) 3n 81 (integer) Thus, the basis step The Induction Step: holds (holds/fails) Assume that 10k <3 is true for some integer k > 4 Then 3k+1 = 3. 3^k 20k 3. 10k > 10k+ where the last inequality is true because k > 4 20k ≥ 80. = 10k+ = 10(k + 1), which means that

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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60 Use Induction to prove that 10n < 3" for all integers n > 4. Proof: The Basis Step: n = 4 10n 40 = (integer) 3n 81 (integer) Thus, the basis step The Induction Step: holds (holds/fails) Assume that 10k <3 is true for some integer k > 4 Then 3k+1 = 3. 3^k 20k 3. 10k > 10k+ where the last inequality is true because k > 4 20k ≥ 80. = 10k+ = 10(k + 1), which means that
Transcribed Image Text:60 Use Induction to prove that 10n < 3" for all integers n > 4. Proof: The Basis Step: n = 4 10n 40 = (integer) 3n 81 (integer) Thus, the basis step The Induction Step: holds (holds/fails) Assume that 10k <3 is true for some integer k > 4 Then 3k+1 = 3. 3^k 20k 3. 10k > 10k+ where the last inequality is true because k > 4 20k ≥ 80. = 10k+ = 10(k + 1), which means that
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