How do you use math induction to prove 3^n > 2n ?
Thank you!
for n = 0
3^0 = 1
2*0 = 0 so true
for n=1
3^1 = 3
2n = 2 so true
for further n
3^(n+1) = 3 * 3^n
2(n+1) = 2 + 2n
for any n bigger than 1 is 3^n > 2 n ??
n = 2
9 > 4
n=3
27 > 6
n =4
81 > 8
see what is happening?
They all worked?
But I don't get the fuurther n part...
for further n
3^(n+1) = 3 * 3^n ( why 3 is multiplied by 3^n)
2(n+1) = 2 + 2n (why is 2 added to3 ^n)
Thank you!
I mean why is 2 added to 2n...?
Oh, I think I got it now
To prove 3^n > 2n using mathematical induction, you need to follow these steps:
Step 1: Base Case
Start by proving the statement for the base case, which is usually n = 1 or n = 0. In this case, let's use n = 1.
When n = 1, we need to show that 3^1 > 2(1). Simplifying this inequality, we get 3 > 2, which is true.
Step 2: Assume the Hypothesis
Now, assume that the statement is true for some positive integer 'k.' This is known as the induction hypothesis.
Hypothesis: Assume that 3^k > 2k.
Step 3: Inductive Step
Now, we need to prove that the statement is true for the next positive integer, which is 'k + 1.' This is called the inductive step.
To do this, we start with the left-hand side of what we're trying to prove, which is 3^(k+1), and we need to show that it is greater than 2(k+1).
Using the induction hypothesis, we know that 3^k > 2k.
Now, let's consider 3^(k+1) = 3 * 3^k. By substituting the induction hypothesis into this equation, we get:
3^(k+1) = 3 * (3^k) > 3 * 2k (since we assumed 3^k > 2k)
Next, we simplify further:
3 * 2k = 6k
Therefore, we can rewrite our inequality as:
3^(k+1) > 6k.
Step 4: Conclusion
By applying the principle of mathematical induction, we have shown that if the statement is true for n = 1 (our base case) and if it is true for k, then it is also true for k + 1.
Hence, by using mathematical induction, we have proven that for all positive integers n, 3^n > 2n.