Use mathematical induction to give a proof that 3^n < (n+2)! for each integer n>=0.
So far I have:
3^k < (k+2)! is true for 0.
For k+1, 3^(k+1) < (k+3).
Kind of at a loss on where to go next.
Thanks for any help!
Ian
Favorite Answer
Let k >= 0.
The final step is to establish the "domino effect", that is, *assuming* 3^k < (k+2)!
and *proving* 3^(k+1) < (k+3)!.
3^(k+1) = 3^k * 3^1
= 3^k * 3
<= 3^k * (k+3) since k >= 0 and 3^k is positive
< (k+2)! * (k+3) since k + 3 > 0 and we're assuming 3^k < (k+2)!
= (k+3)!.
This completes the proof.
Lord bless you today!
?
Test the claim for n = 0.
3^0 = 1 < 2 = (0 + 2)!
Next, suppose that the claim is true for n = k, that is
3^k < (k + 2)!
We'll try to prove that the claim is true for n = k + 1
3^(k + 1) = 3 * 3^k
< 3 * (k + 2)!
<_ (k + 3)(k + 2)! = (k + 3)!
And we're done.