Thread: [SOLVED] Help agent!

1. [SOLVED] Help agent!

I can't solve this exercise,please help me.
Calculate,
S=1!+2!+3!..............+2006!

2. I searched some math sites. It seems to me that there is no elegant solution. Your problem is based on something called the "left factorial".
$n\geq 1$ we define,
$!n=(n-1)!+(n-2)!+...+2!+1!$
Then,
$!n=\left[ \frac{n!}{e} \right]$
Where, $[ \, ]$ is the nearest interger function.
Thus,
$1!+...+2006!=!2007$
And thus,
$1!+...+2006!=\left[ \frac{2007!}{e} \right]$

3. Hi Hckr:

I don't get it. 4!/e rounds to 9, and 1! +2! + 3! = 9. Yea!

1! + 2! = 3, but 3! / e rounds to 2.

1! + 2! + 3! + 4! = 33, but 5!/e rounds to 44.

Thanks in advance,

Rich B.

4. A different approach...

I'm not entirely sure if this shortens anything...

We have a number $x$ such that $x=1!+2!...2005!+n!$ it can be rewritten as
$x=1+2(1(n-1)+$ $3(n-2)+4(n-3)....2005(n-2004)+2006(n-2005))$

I can simplify further, but I'll let you guys decide the best method...