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
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 27E
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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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