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n^2 < 2^n for all natural numbers n \geq 5.

Answers (1)

P(n)=n^2 < 2^n for  n \geq 5.....(given)

Let us consider

P(k) = k^2< 2^k to be true,

Thus, P(k+1) = (k+1)^2

                     = k^2 + 2k + 1

2^{k+1} = 2(2^k) > 2k^2

Now, since, n^2> 2n + 1 for n\geq 3

k^2 + 2k + 1 < 2k^2

Thus, (k+1)^2< 2^{k+1}

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n2< 2n it is true that, P(n)=n^2 < 2^n for  n \geq 5..

Posted by

infoexpert21

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