(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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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?
Sorry, (n+2)!/n!=56
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?
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
Last edited by mr fantastic; Dec 9th 2008 at 12:54 AM. Reason: Deleted link to commercial website
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 $
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?
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 $
Originally Posted by chiph588@ 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 $ Small correction: n = -9, 6. But only n = 6 is a valid solution to the original equation (to the OP: why?).
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