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Question
Prove the statement by using the Principle of Mathematical Induction:
n2 < 2n for all natural numbers n ≥ 5.
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Solution
P(n) is n2 < 2n for n ≥ 5.
Let P(k) = k2 < 2k be true.
⇒ P(k + 1) = (k + 1)2
= k2 + 2k + 1
2k+1 = 2(2k) > 2k2
Since, n2 > 2n + 1 for n ≥ 3.
We get that,
k2 + 2k + 1 < 2k2
⇒ (k + 1)2 < 2(k+1)
⇒ P(k + 1) is true when P(k) is true.
Therefore, by Mathematical Induction, P(n) = n2 < 2n is true for all natural numbers n ≥ 5.
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