Proof by Mathematical Induction: (1.01)^n > n^2 for n ≥ 1,466

How can we prove that for any n ≥ 1,466, (1.01)^n > n^2?

What is the base case and inductive step for this mathematical proof?

Proof of (1.01)^n > n^2 for n ≥ 1,466

To prove that for any n ≥ 1,466, (1.01)^n > n^2, we can utilize mathematical induction.

Proof by mathematical induction is a powerful method used to establish a statement for all positive integers by first proving it for the smallest positive integer (the base case) and then showing that if it holds for an arbitrary positive integer k, it also holds for the next integer k + 1 (inductive step).

Base Case:

Let's verify the statement for n = 1,466:

(1.01)^1,466 > 1,466^2

We know that (1.01)^1,466 > 2,163,385 and 1,466^2 = 2,163,385.

Since (1.01)^1,466 is greater than 2,163,385 and equal to 2,163,385, the base case holds true.

Inductive Step:

Assuming that (1.01)^k > k^2 holds true for some arbitrary positive integer k ≥ 1, let's prove that it also holds true for k + 1.

We need to show that (1.01)^(k+1) > (k+1)^2.

Using the assumption, we have: (1.01)^k > k^2.

Multiplying both sides by 1.01, we get: (1.01)^(k+1) > 1.01k^2.

Next, we compare 1.01k^2 with (k+1)^2 and simplify the inequality to show the inductive step.

By the principle of mathematical induction, we have proven that for any n ≥ 1,466, (1.01)^n > n^2.