proof by mathematical induction
posted by Cait .
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)(21)
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!!
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