x! - [(x-1)!+(x-2)!+(x-3)!...............2!+1!]

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- Nov 30th 2011, 08:50 AMlivinggourmandsimplify (factorial expression)
x! - [(x-1)!+(x-2)!+(x-3)!...............2!+1!]

- Dec 1st 2011, 06:14 AMAnnatalaRe: simplify (factorial expression)
Intuitive guess, based on 5:

expression is 5*4*3*2*1 - 4*3*2*1 - 3*2*1 - 2*1 - 1 = 4*4*3*2*1 - 3*2*1 - 2*1 - 1 = 4*3*2*2*1 etc.

So I think the answer is going to be (x-1)!, though that's not a formal proof. - Dec 1st 2011, 06:34 AMPlatoRe: simplify (factorial expression)
$\displaystyle 5! - \left( {\sum\limits_{k = 1}^4 {(5 - k)!} } \right) = 87$.

If $\displaystyle E_N=N! - \left( {\sum\limits_{k = 1}^{N-1} {(N - k)!} } \right)$ for $\displaystyle N\ge 2$ then we get $\displaystyle 1,~3,~15,~87,~567$ for the first five of those.

I have not seen a pattern yet. - Dec 1st 2011, 09:48 AMemakarovRe: simplify (factorial expression)
See also the MathWorld article on the sum of factorials.

- Dec 1st 2011, 09:53 AMPlatoRe: simplify (factorial expression)
Also looking at this sequence site we see that this does not to yield much.

- Dec 1st 2011, 04:03 PMlivinggourmandRe: simplify (factorial expression)
do v have any other expression for sum of factorial ??

may be we can find a range (in terms of n) in which the answer would lie..

I ll try this now... - Dec 1st 2011, 04:56 PMPlatoRe: simplify (factorial expression)
- Dec 2nd 2011, 08:54 AMAnnatalaRe: simplify (factorial expression)
Ah, I figured out the mistake in my quickie guess-calculation.

a * b * c - b * c = (a-1) * b * c is true, but a * b * c - c = (a * (b-1) * c) is false. It should = ((a * b) - 1) * c, which doesn't simplify and gets worse the longer the expression is.