Question

Prove by Principle of Mathematical Induction that for every positive integers
n , 3^n >n^2

Answer 1

you haven't really proved that
k^2 * 3 > k^2 + k(k+2) + 1
But working it a bit, we get
3k^2 > k^2 + k^2 + 2k + 1
3k^2 > 2k^2 + 2k + 1
k^2 > 2k + 1
k^2 - 2k + 1 > 0
(k-1)^2 > 0
This is in fact true for all k > 1

Answer 2

Base Case: When n = 1, 3^1 > 1^2 is true since 3 > 1.

Inductive Step: Assume that for some positive integer k, 3^k > k^2 is true.

We need to prove that 3^(k+1) > (k+1)^2 is true.

Starting with the left-hand side, we have:

3^(k+1) = 3^k * 3

Using the assumption, we can replace 3^k with k^2 to get:

3^(k+1) = k^2 * 3

Now, let's simplify the right-hand side:

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

Since we already have k^2 on the left-hand side, we'll focus on the 2k and 1 terms. We can start by factoring out k^2 from the first two terms:

k^2 + 2k = k(k+2)

Now we can substitute that into the right-hand side:

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

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

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

Now we can compare the left-hand side and right-hand side:

3^(k+1) = k^2 * 3 > k^2 + k(k+2) + 1 = (k+1)^2

Therefore, 3^(k+1) > (k+1)^2 is true.

By the principle of mathematical induction, we have proven the statement for all positive integers n.

Answer 3

You're right, thank you for catching that error in my proof! And good job on providing the correct steps for the inequality. The proof is complete once we establish that (k-1)^2 > 0 is true for all k > 1.