1. ## factorial 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?

2. you are saying to divide $n+2$ by $n!$ and get 56.

This can never happen though...
Are you sure you have the right problem?

3. Sorry,

(n+2)!/n!=56

4. Note that $(n+2)! = (n+2)(n+1)(n)(n-1)(n-2) \dots = (n+2)(n+1)n!$

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

Do you see where to go now?

5. 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

6. You solved that wrong...

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

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

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

$n=-6,9$

7. 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?

8. its simple factoring

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

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

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

By inspection it can be seen $a=-9 , \; b=6$

9. Originally Posted by chiph588@
You solved that wrong...

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

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

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

$n=-6,9$
Small correction: n = -9, 6.

But only n = 6 is a valid solution to the original equation (to the OP: why?).