[ Algebra and arithmetic (miscellaneous).] $[$ Sequences (miscellaneous) ]

Find the value of the expression $1!* 3-2!* 4+3!* 5-4!* 6+\ldots-2000!* 2002+2001!$.
Start simplifying the given expression from the end.

## Solution

Let's denote the given expression by S and start transforming it from the end: -2000!*2002 + 2001! = 2000! * (-2002 + 2001) = -2000!. Thus, S = 1!*3 - 2!*4 + 3!*5 - 4!*6 + ... + 1999!*2001 - 2000!. Next, 1999!*2001 - 2000! = 1999!(2001 - 2000) = 1999!. Therefore, S = 1!*3 - 2!*4 + 3!*5 - 4!*6 + ... - 1998!*2000 + 1999!. By continuing to simplify the sum S in this manner, we will arrive at S = 1! = 1.