# Thread: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)

1. ## 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)

Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

thanks

2. Originally Posted by rcs

Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

thanks
id:A033312 - OEIS Search Results

3. Originally Posted by rcs

Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

thanks
Clearly:

1! = 2! - 1.

1(1!)+ 2(2!) = 3! - 1.

1(1!)+ 2(2!) + 3(3!) = 4! -1.

There is a clear pattern that suggests a hypothesis for the general formula. It is not difficult to prove the hypothesis true using induction.

4. proof without induction: since $\displaystyle k \cdot k! = (k+1)! - k!$, we'll have a telescoping sum:

$\displaystyle \sum_{k=1}^n k \cdot k! = \sum_{k=1}^n [(k+1)! - k!]=(n+1)!-1!=(n+1)!-1.$