Thread: Define n!

im trying to answer or show the proof but unfortunately there things that are not clear to me

Problem:

Find the sum of the series 1 (1!) + 2 (2!) + 3 (3!) + ... + n (n!)
Define n! = n (n - 1) (n - 2) ( n - 3) ... (3)(2)(1)
Example 8! = 8(7)(6)(5)(4)(3)(2)(1)

My attempt:

=> 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!) Row 1
add 1 => 2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + n - 1+ 1(n-1)! + (n+1)(n!) Row 2
Subtract (Row 2 - Row 1)

2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + (n-1+ 1)(n-1)! + (n+1)(n!)
- 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!)
1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + ... 1 (n -1)! +n (n!) Row 3 ( answer)

Subtract the values of Row 2 and Row 3, then
2 (1!) + 3(2!) + 4(3!) + 5 (4!)+ ... + (n - 1+ 1)(n-1)! + (n+1)(n!) Row 2
- [ 1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + 1(5!)+ ... 1 (n -1)! + n (n!)


therefore
(n+1)(n!) - 1 (1!)
(n+1)! - 1 this is my answer


Please correct me if im wrong

Thanks

Originally Posted by rcs

im trying to answer or show the proof but unfortunately there things that are not clear to me

Problem:

Find the sum of the series 1 (1!) + 2 (2!) + 3 (3!) + ... + n (n!)
Define n! = n (n - 1) (n - 2) ( n - 3) ... (3)(2)(1)
Example 8! = 8(7)(6)(5)(4)(3)(2)(1)

My attempt:

=> 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!) Row 1
add 1 => 2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + n - 1+ 1(n-1)! + (n+1)(n!) Row 2
Subtract (Row 2 - Row 1)

2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + (n-1+ 1)(n-1)! + (n+1)(n!)
- 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!)
1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + ... 1 (n -1)! +n (n!) Row 3 ( answer)

Subtract the values of Row 2 and Row 3, then
2 (1!) + 3(2!) + 4(3!) + 5 (4!)+ ... + (n - 1+ 1)(n-1)! + (n+1)(n!) Row 2
- [ 1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + 1(5!)+ ... 1 (n -1)! + n (n!)


therefore
(n+1)(n!) - 1 (1!)
(n+1)! - 1 this is my answer


Please correct me if im wrong

Thanks

The way you have written this has a lot to be desired, but your answer appears correct...

Originally Posted by rcs

Problem:
Find the sum of the series 1 (1!) + 2 (2!) + 3 (3!) + ... + n (n!)
Define n! = n (n - 1) (n - 2) ( n - 3) ... (3)(2)(1)
Example 8! = 8(7)(6)(5)(4)(3)(2)(1)
My attempt:
=> 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!) Row 1
add 1 => 2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + n - 1+ 1(n-1)! + (n+1)(n!) Row 2
Subtract (Row 2 - Row 1)
2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + (n-1+ 1)(n-1)! + (n+1)(n!)
- 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!)
1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + ... 1 (n -1)! +n (n!) Row 3 ( answer)
Subtract the values of Row 2 and Row 3, then
2 (1!) + 3(2!) + 4(3!) + 5 (4!)+ ... + (n - 1+ 1)(n-1)! + (n+1)(n!) Row 2
- [ 1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + 1(5!)+ ... 1 (n -1)! + n (n!)

therefore
(n+1)(n!) - 1 (1!)
(n+1)! - 1 this is my answer CORRECT

You have the correct answer, but frankly I cannot follow your work.

This is quite easy to show using induction.

Originally Posted by Prove It

The way you have written this has a lot to be desired, but your answer appears correct...

Thank you my IDOL sir ProveIT. you never put me down. You are the BEST sir