Prove using mathematical induction for n = 5: 2^n n^2.

Question

Prove using mathematical induction for n >= 5: 2^n > n^2.

Accepted Answer

By mathematical induction, 2^n > n^2 holds for all integers n >= 5. We want to show 2^n grows faster than n^2 for large n. Mathematical induction works perfectly here: verify the base case at n = 5, then show that if the inequality holds at n = k it must also hold at n = k+1. The key insight for the inductive step is that doubling 2^k (multiplying by 2) beats the needed growth from k^2 to (k+1)^2 once k >= 5.