# Thread:

1. ## Prove By Mathematical Induction

I'm stuck on this question:

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

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

So far I've done this:

$P(n)$ is a statement for each natural number $n$, where $n \geq 4$. $P(n)$ 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 $n = 4$, $4(4!) > 2^{6}$, $96 > 64$ hence true.

For $n + 1$:

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

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

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

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

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

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

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

So far I've done this:

$P(n)$ is a statement for each natural number $n$, where $n \geq 4$. $P(n)$ 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 $n = 4$, $4(4!) > 2^{6}$, $96 > 64$ hence true.

For $n + 1$:

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

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

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

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

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

P(k)

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

P(k+1)

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

Proof

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

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

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

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

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

3. That's great, thanks
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