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Here is a question and solution but I don't understand what's happening after $m = m+1$.

How does $(3(m+1))!$ equal $(3m)!(3m+1)(3m+2)(3m+3)$? Should it not be $(3m+3)!$?

Same thing with the other side of the equation.

I also don't understand the equation after the "Also".

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$(3m+3)!=(3m+3)(3m+2)(3m+1)(3m)!$ Trying playing around with some factorials a little so you can get comfortable with them. –  Joseph G. Apr 30 '13 at 12:50
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For after the also they are just saying that $3m+3 \ge 3m+2$ and $3m+3 \ge 3m+1$ so then we can say that $(3m+3)(3m+3)(3m+3) = (3m+3)^3 \ge (3m+1)(3m+2)(3m+3)$ because the first two terms are large on the left than the right side. Try it for $m=0$ if you cant understand what they are doing. Then $(3m+3)^3 = (3*(m+1))^3=3^3*(m+1)^3=27(m+1)^3$ for the last equality. –  Ben Apr 30 '13 at 12:56
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1 Answer

up vote 3 down vote accepted

$$(3(m+1))!=(3m+3)!=1\cdot2\cdot\ldots\cdot(3m)\cdot(3m+1)\cdot(3m+2)\cdot(3m+3)=$$

$$=(3m)!(3m+1)(3m+2)(3m+3)$$

And a similar thing on the other side.

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Thanks. What about this part? $(3m + 1)(3m + 2)(3m + 3)≤(3m + 3)^3 = 27(m + 1)^3$ –  Adegoke A Apr 30 '13 at 12:54
    
Isn't it pretty straightforward that $\,3m+1\,,\,3m+2\le 3m+3\,$ ? So each of the three factors on the left is less than or equal to the rightmost one of them... –  DonAntonio Apr 30 '13 at 12:55
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