I am trying to prove the equation for any integer (k) greater or equal to 0, that
2^(k+1) < (k+3)!
I am not sure how to expand the factorial to show that the equation will always be true?
Any Help Would Be Greatly Appreciated
If we write...
$\displaystyle 2^{k+1}= \prod_{i=1}^{k+1} a_{i}$ (1)
$\displaystyle (k+3)!= \prod_{i=1}^{k+2} b_{i}$ (2)
... You observe that the product (2) has k+2 factors and the product (1) k+1 factors and for i=1,2,...,k+1 is $\displaystyle a_{i}\le b_{i}$, so that the conclusion is obvious...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
A more elementary concept probably doesn't exist...
$\displaystyle 2^{k+1}= 2 \cdot 2 \cdot ... \cdot 2\ (k+1\ \text{times})$ (1)
$\displaystyle (k+3)!= 2 \cdot 3 \cdot ... \cdot (k+2)\ \cdot (k+3)$ (2)
Do You understand now?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
I understand how to solve each side of the equation, but what I am confused about is how did you determine the factors in your first response, and then how do I compare those two products to show that the one will always be larger than the other?
Hello, doleary22!
Here is an observation.
I hope you can modify it into a proof.
$\displaystyle \text{Prove that, for any integer }k \ge 0\!:\;\;2^{k+1} \:<\: (k+3)!$
$\displaystyle \text{We have: }\;\underbrace{2\cdot2\cdot2\,\cdots\,2} \quad ^<_>\quad 1\cdot2\cdot\underbrace{3\cdot4\,\cdots\,(k+3)}$
. . . . . . . . . $\displaystyle ^{k+1\text{ factors}} \qquad\qquad\qquad \quad ^{k+1\text{ factors } >\: 2} $