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Prove by mathematical induction that n^2 +n < 2^n whenever n is an integer greater than 4.

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Let P(n) be the proposition that n^2+n< 2^n, n> 4.

Basis Step

P(5) is true, because 5^2+5=30<32=2^5.

Inductive Step

We assume that P(k) is true for an arbitrary integer k with k > 4. That is, we assume that k^2+k< 2^k, for the positive integer k with k > 4.

We must show that under this hypothesis, P(k + 1) is also true. That is, we must show that if k^2+k< 2^k for an arbitrary positive integer k where k > 4, then

(k+1)^2+k+1< 2^{k+1}, k>4

We have for k > 4


2^{k+1}=2\cdot2^k>2(k^2+k)=k^2+2k+k^2

And

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

This shows that P(k + 1) is true when P(k) is true. This completes the inductive step of the proof.

We have completed the basis step and the inductive step. Hence, by mathematical induction P(n) is true for all integers n with n > 4. That is, we have proved that n^2+n< 2^n is true for all integers n with n > 4.

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