Use math induction to show that

1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6

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- Mar 15th 2008, 09:29 PMxomichelleybellyMath Induction...PLEASE HELP!!!!!
Use math induction to show that

1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6 - Mar 16th 2008, 12:24 AMCaptainBlack
I will assume that you have done the base case and that we assume this is true for some $\displaystyle k \ge 6$

So we need to show that:

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

well:

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

and by supposition $\displaystyle k! \ge k^2$, so:

$\displaystyle (k+1)! = k! k \ge k^3$

So if we can prove that $\displaystyle k^3 \ge (k+1)^2$ we are done, but $\displaystyle k^3 \ge (k+1)^2$ is equvalent to $\displaystyle k^3-k^2-2k-1 \ge 0$. So to compete the proof of the induction step you need to show that:

$\displaystyle k^3-k^2-2k-1 = 0$

has no real root greater than or equal to $\displaystyle 6$ (this is sufficient as for $\displaystyle k$ greater than its largest real root the cubic is positive).

This you do by making the substitution $\displaystyle x=k-6$, then using Descartes rule of signs to show that this has no positive roots.

(In fact the only real root of $\displaystyle k^3-k^2-2k-1 = 0$ is near $\displaystyle k=2.1$)

RonL - Mar 16th 2008, 12:54 AMCaptainBlack
- Mar 16th 2008, 01:01 AMmr fantastic