Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - Define n!

  1. #1
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,589
    Thanks
    1445

    Re: Define n!

    Quote Originally Posted by rcs View Post
    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*}
    Thanks from rcs
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Define n!

    Quote Originally Posted by rcs View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2

    Re: Define n!

    Quote Originally Posted by Prove It View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Define each of the following.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 8th 2013, 03:19 PM
  2. define x ~ y
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 30th 2011, 01:10 PM
  3. [SOLVED] {a_n}\to A and define {b_n}
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: June 19th 2011, 07:12 PM
  4. Replies: 1
    Last Post: March 13th 2010, 12:13 PM
  5. Define gcd(a,b,c)
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 18th 2009, 06:00 AM

Search Tags


/mathhelpforum @mathhelpforum