Sequences - factorial r, !

[martinrandau]

Can anybody help me solving this?

Write in terms of factorials

n((n^2)-1)

The correct answer is
(n+1)!/(n-2)!

but I don't know how to get there, and since it's week- end I have no chance to ask anyone teachers, etc.
//Martin

[Guybrush Threepwood]

n(n2-1) = n(n-1)(n+1)

[martinrandau]

[QUOTE]Originally posted by martinrandau
[B]Can anybody help me solving this?

Write in terms of factorials

n((n^2)-1)

The correct answer is
(n+1)!/(n-2)!


Please notice the expression marks (!). The task is not to factorise it by "normal" means, but to find an expression as a sequence.
ex. 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
n!= 1 x 2 x 3 x...x n

It's called the factorial r (!).
Thank you for your help anyway!

[Tom Mattson]

Originally posted by martinrandau
The correct answer is
(n+1)!/(n-2)!


I'll give you a hint.

Expand the numerator and denominator of the above ratio and cancel the factors common to both. For instance, the numerator is:

(n+1)!=(n+1)(n)(n-1)(n-2)...

Get the idea?

[martinrandau]

Yes![:)]
Thank you!

//Martin