induction inequalities P2

**christina** · February 13th 2010, 03:39 AM

Prove using the principle of mathematical induction:

n!>n^2 for n=4,5,6,7...

thank you so much!

**Krizalid** · February 13th 2010, 06:33 AM

show your progress.

**Roam** · February 13th 2010, 10:51 AM

Originally Posted by christina

Prove using the principle of mathematical induction:

n!>n^2 for n=4,5,6,7...

thank you so much!

Read your post again. You are not asking any question or asking for help. You are telling us to do something. Do you even know the axiom of induction? Let P(n) be your proposition about the natural number n. Then you have to prove two things (1) The base case, $P(1)$ is true; $1! \geq 1^2$ . Now comes the hard part, for each $k \in N$ , whenever $P(k)$ is true, then $P(k+1)$ is true. So you know that

$1 \times ... \times k \geq k^2 = k \times k$

$LHS= (1 \times ... \times k) \times (k+1) \geq k^2 \times (k+1)$ (by the inductive hypothesis)

Can you complete the induction?

**Archie Meade** · February 13th 2010, 05:14 PM

Originally Posted by christina

Prove using the principle of mathematical induction:

n!>n^2 for n=4,5,6,7...

thank you so much!

$n!>n^2,\ n\ge4$

Therefore, if that's true, so should the following

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

Attempt to prove this using $n! > n^2$

so that true for n causes true for n+1

$(k+1)k!>(k+1)k^2$ ?

$(k+1)!>(k+1)k^2$ ?

$(k+1)!>k^2+k^3$ ?

$(k+1)!>k^2+k^2$ ?

$(k+1)!>k^2+4k,\ k \ge 4$

$(k+1)!>k^2+2k+1$

$(k+1)!>(k+1)^2$

Test for n=4

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