# Thread: [SOLVED] Prove this with induction

1. ## [SOLVED] Prove this with induction

$\displaystyle 3.2^n<n!$ for any positive interger n greater than 4

i did the base step

$\displaystyle 3\times2^5<5!$
$\displaystyle 96<120$

now the n+1 step, i can do some of but get stuck

$\displaystyle 3\times2^{n+1}<(n+1)!$
$\displaystyle 3\times2^n\times2<n!\times(n+1)$

what do i do from here? thanks.

2. Originally Posted by Rapid_W
$\displaystyle 3.2^n<n!$ for any positive interger n greater than 4

i did the base step

$\displaystyle 3\times2^5<5!$
$\displaystyle 96<120$
At this point you assume that $\displaystyle 3\cdot 2^n <n!$ is a true statement.
Next we take a look at $\displaystyle 3\cdot 2^{n+1}$

But that is $\displaystyle 2\cdot 3\cdot 2^n <2n!$ from the inductive assumption.
How does 2 compare with n+1? Now you have one more step.

3. Did you get 2n! because all you have done to the LHS is essentially multiply by two?

Also is the next step just stating that 2<n+1 when n<4?

4. Originally Posted by Rapid_W
Did you get 2n! because all you have done to the LHS is essentially multiply by two?
Also is the next step just stating that 2<n+1 when n<4?
We know that $\displaystyle \color{blue}3\cdot 2^n<n!$
Looking at $\displaystyle 3\cdot 2^{n+1}=2\cdot{\color{blue}3\cdot 2^n}<2\cdot\color{blue}n!$.

Because $\displaystyle 2 < 4 < n < n + 1$ and $\displaystyle (n+1)n!=(n+1)!$ you are done.