I can't solve this exercise,please help me.
Calculate,
S=1!+2!+3!..............+2006!
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".
$\displaystyle n\geq 1$ we define,
$\displaystyle !n=(n-1)!+(n-2)!+...+2!+1!$
Then,
$\displaystyle !n=\left[ \frac{n!}{e} \right]$
Where, $\displaystyle [ \, ]$ is the nearest interger function.
Thus,
$\displaystyle 1!+...+2006!=!2007$
And thus,
$\displaystyle 1!+...+2006!=\left[ \frac{2007!}{e} \right]$
I'm not entirely sure if this shortens anything...
We have a number$\displaystyle x$ such that $\displaystyle x=1!+2!...2005!+n!$ it can be rewritten as
$\displaystyle x=1+2(1(n-1)+$$\displaystyle 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...