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Using the principle of mathematical induction, prove each of the following for all n ϵ N: 

3n ≥ 2n .

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To Prove:

3n ≥ 2n

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n): 3n ≥ 2n

For n = 1 P(n) is true since 3n ≥ 2n i x e x 3 ≥ 2, which is true

Assume P(k) is true for some positive integer k , ie,

= 3k ≥ 2k....(1)

We will now prove that P(k + 1) is true whenever P( k ) is true 


[Multiplying and dividing by 2 on RHS]

Therefore, P (k + 1) is true whenever P(k) is true 

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N. 

Hence proved.

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