(n+2)/n!=56

I'm new to this and having trouble trouble breaking this down, could someone please give me a hint to get this started?

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- Dec 8th 2008, 09:24 PMCarrickfactorial notation help
(n+2)/n!=56

I'm new to this and having trouble trouble breaking this down, could someone please give me a hint to get this started? - Dec 8th 2008, 09:29 PMchiph588@
you are saying to divide $\displaystyle n+2 $ by $\displaystyle n! $ and get 56.

This can never happen though...

Are you sure you have the right problem? - Dec 8th 2008, 09:33 PMCarrick
Sorry,

(n+2)!/n!=56 - Dec 8th 2008, 09:38 PMchiph588@
Note that $\displaystyle (n+2)! = (n+2)(n+1)(n)(n-1)(n-2) \dots = (n+2)(n+1)n! $

So $\displaystyle \frac{(n+2)!}{n!} = \frac{(n+2)(n+1)n!}{n!} $

Do you see where to go now? - Dec 8th 2008, 09:42 PMwinsome
Given: (n + 2)!/n! = 56

(n + 2)(n + 2 - 1)(n + 2 - 2)!/n! = 56

(n + 2)(n + 1) = 56

n + 2 = 56 or n + 1 = 56

Which gives n = 54, 55

May be it helps You - Dec 8th 2008, 09:49 PMchiph588@
You solved that wrong...

$\displaystyle (n+2)(n+1) = n^2+3n+2=56 $

So $\displaystyle n^2+3n-54 = 0 $

$\displaystyle (n+9)(n-6) = 0 $

$\displaystyle n=-6,9 $ - Dec 8th 2008, 09:56 PMCarrick
thanks for the help chip, when down to

(n+9)(n-6) how do you know which to make positive and negative?

eg how do you know its not (n+6)(n-9) without substiting into the initial equation? - Dec 8th 2008, 10:06 PMchiph588@
its simple factoring

given that $\displaystyle (n-a)(n-b) = n^2+3n-54 $

doing foil... $\displaystyle n^2+(-a-b)n+ab = n^2+3n-54 $

So $\displaystyle a+b = -3 $ and $\displaystyle ab = -54 $

By inspection it can be seen $\displaystyle a=-9 , \; b=6 $ - Dec 9th 2008, 01:16 AMmr fantastic