College

Prove that for any \( n \geq 1,466 \), [tex](1.01)^n > n^2[/tex].

You may use the fact that [tex](1.01)^{1,466} > 2,163,385[/tex] and [tex]1,466^2 = 2,163,385[/tex].

Answer :

To prove that for any n ≥ 1,466, (1.01)ⁿ > n², we can use mathematical induction.

Base Case:

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

(1.01)¹,466 > 1,466²

We are given that (1.01)¹,466 > 2,163,385 and 1,466² = 2,163,385.

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

Inductive Step:

Assuming that (1.01)^k > k² 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 > k²

Multiplying both sides by 1.01, we get:

(1.01)^k * 1.01 > k² * 1.01

Simplifying:

(1.01)^(k+1) > 1.01k²

Now, we want to compare 1.01k² with (k+1)²:

1.01k² < (k+1)²

Expanding (k+1)²:

1.01k² < k² + 2k + 1

Since k ≥ 1, we know that k² > 0 and 2k > 0.

Therefore, 1.01k² < k² + 2k + 1 holds true.

Combining this inequality with our previous result:

(1.01)^(k+1) > 1.01k² > k² + 2k + 1 = (k+1)²

Thus, if (1.01)^k > k², then (1.01)^(k+1) > (k+1)².

By the principle of mathematical induction, we have proven that for any n ≥ 1,466, (1.01)ⁿ > n².

Learn more about mathematical induction:

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