subject is PreCalulus.
2^(k+3) = and < (k+3)!
i know how to do proving using mathematical induction when its just an equal sign, but I don't understand what to do when its an inequality.
thank you!!!
Here is what I would do
Step 1: check if it is true for k=1
that is: let k=1
2^4<= 4! ?
16<=24 ? YES
Step 2: assume it is true for k=n
or 2^(n+3) <= (n+3)! is true
Step 3: test to see if it is then true for k= n+1
or, is 2^(n+4) <= (n+4)! ??
subtract the two inequations (step 4 minus step3)
is 2^(n+4) - 2^(n+3) <= (n+4)! - (n+3)! ???
the left side factors to
2^(n+3)(2-1)
or 2^(n+3)
the right side is (n+4)*(n+3)! .... {just like 8! - 7! = 8*7!}
so we are looking at:
is 2^(n+3) <= (n+4)*(n+3)! ??
Well, the left side is the original left side of step 2, and the right side is the original right side of step 2 multiplied by a positive number > 1.
Since the right side was already >= the left side, muliplying it by a positive integer would cerainly make it even larger.
thank you lots!!
3467800
To prove the inequality 2^(k+3) < (k+3)!, you can use mathematical induction. Here's a step-by-step explanation of how to do it:
Step 1: Check the base case.
Let k = 1 and substitute it into the inequality:
2^(1+3) < (1+3)!
16 < 24
Since this is true, we move on to the next step.
Step 2: Assume the inequality is true for k = n.
Assume that 2^(n+3) < (n+3)! is true. This is called the induction hypothesis.
Step 3: Prove that the inequality holds for k = n+1.
Now, we need to prove that if 2^(n+3) < (n+3)!, then 2^(n+4) < (n+4)!.
Step 4: Subtract the inequality for k = n+1 from the inequality for k = n.
Subtract the two inequalities:
2^(n+4) - 2^(n+3) < (n+4)! - (n+3)!
Step 5: Simplify both sides of the inequality.
The left side simplifies to 2^(n+3)(2-1) = 2^(n+3).
The right side simplifies to (n+4)(n+3)!.
Step 6: Compare the values on both sides of the inequality.
We end up with the inequality 2^(n+3) < (n+4)(n+3)!.
Step 7: Use the induction hypothesis.
Since the induction hypothesis states that 2^(n+3) < (n+3)!, and (n+4) is a positive integer greater than 1, multiplying it by (n+3)! would certainly make it larger.
Step 8: Conclusion.
Therefore, using the principle of mathematical induction, we can conclude that 2^(k+3) < (k+3)! holds for all positive integers k.
If you have any further questions or need clarification, feel free to ask!