# Define n!

• Aug 9th 2012, 12:28 AM
rcs
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
• Aug 9th 2012, 06:07 AM
Prove It
Re: Define n!
Quote:

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...

\displaystyle \begin{align*} S &= \sum_{k = 1}^{n}{\left(k\cdot k!\right)} \\ &= \sum_{k = 1}^n{\left(k\cdot k! + k! - k!\right) } \\ &= \sum_{k = 1}^{n}{\left[(k + 1)\cdot k! - k!\right]} \\ &= \sum_{k = 1}^n{\left[(k + 1)\cdot k!\right]} - \sum_{k = 1}^{n}\left(k!\right) \\ &= \sum_{k = 1}^n{\left[(k + 1)!\right]} - \sum_{k = 1}^n{\left(k!\right)} \\ &= \left[2! + 3! + 4! + \dots + n! + (n+1)!\right] - \left(1! + 2! + 3! + \dots + n!\right) \\ &= (n + 1)! - 1! \\ &= (n + 1)! - 1 \end{align*}
• Aug 9th 2012, 06:10 AM
Plato
Re: Define n!
Quote:

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

This is quite easy to show using induction.
• Aug 10th 2012, 04:24 AM
rcs
Re: Define n!
Quote:

Originally Posted by Prove It
The way you have written this has a lot to be desired, but your answer appears correct...

\displaystyle \begin{align*} S &= \sum_{k = 1}^{n}{\left(k\cdot k!\right)} \\ &= \sum_{k = 1}^n{\left(k\cdot k! + k! - k!\right) } \\ &= \sum_{k = 1}^{n}{\left[(k + 1)\cdot k! - k!\right]} \\ &= \sum_{k = 1}^n{\left[(k + 1)\cdot k!\right]} - \sum_{k = 1}^{n}\left(k!\right) \\ &= \sum_{k = 1}^n{\left[(k + 1)!\right]} - \sum_{k = 1}^n{\left(k!\right)} \\ &= \left[2! + 3! + 4! + \dots + n! + (n+1)!\right] - \left(1! + 2! + 3! + \dots + n!\right) \\ &= (n + 1)! - 1! \\ &= (n + 1)! - 1 \end{align*}

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