# Thread: Prove By Mathematical Induction

1. ## Prove By Mathematical Induction

I'm stuck on this question:

State the principle of mathematical induction, and use it to prove that for $\displaystyle n \geq 4$

$\displaystyle 4n! > 2^{n+2}$

So far I've done this:

$\displaystyle P(n)$ is a statement for each natural number $\displaystyle n$, where $\displaystyle n \geq 4$. $\displaystyle P(n)$ is true if:

• P(n) is true for n = 4.
• If it is true for arbitrary n, it is also true for n + 1.

For $\displaystyle n = 4$, $\displaystyle 4(4!) > 2^{6}$, $\displaystyle 96 > 64$ hence true.

For $\displaystyle n + 1$:

$\displaystyle 4(n + 1)! > 2^{n + 3}$

$\displaystyle (n + 1)! > 2^{n + 1}$

$\displaystyle n!(n+1) > 2^{n}(n+1)$ (Using hypothesis)

And that's where I get stuck. RHS should be $\displaystyle 2^{n+1}$, but that only occurs when n = 1.

Please help!

2. Originally Posted by splbooth
I'm stuck on this question:

State the principle of mathematical induction, and use it to prove that for $\displaystyle n \geq 4$

$\displaystyle 4n! > 2^{n+2}$

So far I've done this:

$\displaystyle P(n)$ is a statement for each natural number $\displaystyle n$, where $\displaystyle n \geq 4$. $\displaystyle P(n)$ is true if:

• P(n) is true for n = 4.
• If it is true for arbitrary n, it is also true for n + 1.

For $\displaystyle n = 4$, $\displaystyle 4(4!) > 2^{6}$, $\displaystyle 96 > 64$ hence true.

For $\displaystyle n + 1$:

$\displaystyle 4(n + 1)! > 2^{n + 3}$

$\displaystyle (n + 1)! > 2^{n + 1}$

$\displaystyle n!(n+1) > 2^{n}(n+1)$ (Using hypothesis)

And that's where I get stuck. RHS should be $\displaystyle 2^{n+1}$, but that only occurs when n = 1.

Please help!
you can try to prove whether or not $\displaystyle 4(n+1)!>2^{(n+1)+2}$ if $\displaystyle 4n!>2^{n+2}$

P(k)

$\displaystyle 4k!>2^{k+2}$

P(k+1)

$\displaystyle 4(k+1)!>2^{(k+1)+2}$

Proof

$\displaystyle 4(k+1)!=4(k+1)k!=(k+1)4k!$

If $\displaystyle k\ge4$ then $\displaystyle (k+1)4k!\ge(5)4k!$

$\displaystyle 2^{k+3}=(2)2^{k+2}$

Then if $\displaystyle 4k!>2^{k+2}$

$\displaystyle 4(k+1)!>2^{k+3}$ certainly since $\displaystyle (5)4k!>(2)2^{k+2}$

3. That's great, thanks