Summation of factorials.

Question

How do I go about summing this :

$$\sum_{r=1}^{n}r\cdot (r+1)!$$

I know how to sum up $r\cdot r!$ But I am not able to do a similar thing with this.

Answer 1

Making use of:

$$r.\left(r+1\right)!=\left(r+2-2\right).\left(r+1\right)!=\left(r+2\right)!-2.\left(r+1\right)!$$

we write the summation as:

$$\left[\left(n+2\right)!-2.\left(n+1\right)!\right]+\left[\left(n+1\right)!-2.n!\right]+\cdots+\left[4!-2.3!\right]+\left[3!-2.2!\right]$$

leading to:

$$\sum_{r=1}^{n}r.\left(r+1\right)!=\left(n+2\right)!-2-\sum_{r=1}^{n}\left(r+1\right)!$$

So finding an expression for it is in essence the same as finding an expression for: $$\sum_{r=1}^{n}r!$$

For this have a look here.