Prove by Principle of Mathematical Induction that for every positive integers
n , 3^n >n^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.
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
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.
To prove that 3^n > n^2 for every positive integer n using the Principle of Mathematical Induction, we will follow these steps:
Step 1: Base Case
First, we will show that the statement is true for the smallest possible value of n. Let's take n = 1.
3^1 = 3, and 1^2 = 1. Since 3 > 1, the statement holds true for n = 1.
Step 2: Inductive Hypothesis
Assume that the statement is true for some positive integer k. That is, assume 3^k > k^2.
Step 3: Inductive Step
We need to prove that the statement is also true for the next positive integer, k + 1. That is, we need to prove that 3^(k+1) > (k+1)^2.
Starting with the left-hand side of the inequality, we have:
3^(k+1) = 3^k * 3
Since we assumed 3^k > k^2 (our inductive hypothesis), we can rewrite this as:
3^(k+1) > k^2 * 3
Now let's simplify the right-hand side of the inequality:
(k+1)^2 = k^2 + 2k + 1
Since k is a positive integer, we know that 2k > 0, and therefore (2k + 1) > 1. Thus, we can write:
(k+1)^2 > k^2 + 1
Combining these results, we have:
3^(k+1) > k^2 * 3 > k^2 + 1 = (k+1)^2
Therefore, we have shown that if the statement is true for some positive integer k, then it is also true for k + 1. This completes the inductive step.
Step 4: Conclusion
By the Principle of Mathematical Induction, the statement holds true for all positive integers n. Therefore, we have proven that 3^n > n^2 for every positive integer n.