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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