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Math Help - Simplified Factorial Notation

  1. #1
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    Smile Simplified Factorial Notation

    Alright.

    So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please?

    So on about factorial notation. I have no idea what it means.

    Key

    _ = sub

    Givens:

    {a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

    C_n = a_n + a_(n + 1)

    I need to write it in factorial notation. And Simplified. Any helpers out there today? =)
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  2. #2
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    Quote Originally Posted by ninjahcamp3r View Post
    Alright.

    So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please?

    So on about factorial notation. I have no idea what it means.

    Key

    _ = sub

    Givens:

    {a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

    C_n = a_n + a_(n + 1)

    I need to write it in factorial notation. And Simplified. Any helpers out there today? =)
    For any n \in N , n! = n*(n-1)*(n-2)...* 1
    So for example,
    5! = 5*4*3*2*1
    and 8! = 8*7*6*5*4*3*2*1

    Hope that helps.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ninjahcamp3r View Post
    Alright.

    So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please?

    So on about factorial notation. I have no idea what it means.

    Key

    _ = sub

    Givens:

    {a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

    C_n = a_n + a_(n + 1)

    I need to write it in factorial notation. And Simplified. Any helpers out there today? =)
    If a_n=n\cdot n!, then it would follow that a_{n+1}=(n+1)\cdot(n+1)!

    So, C_n=a_n+a_{n+1}=n\cdot n!+(n+1)\cdot(n+1)!=\left[n+(n+1)^2\right]n!=\color{red}\boxed{(n^2+3n+1)\cdot n!}

    Does this make sense?
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    If a_n=n\cdot n!, then it would follow that a_{n+1}=(n+1)\cdot(n+1)!

    So, C_n=a_n+a_{n+1}=n\cdot n!+(n+1)\cdot(n+1)!=\left[n+(n+1)^2\right]n!=\color{red}\boxed{(n^2+3n+1)\cdot n!}

    Does this make sense?
    Alright the last part is also what I have but is it all the way simplified.

    I also forgot to include that a_n can also be expressed as (n + 1)! - n!.

    Could you also give me a link on how you guys right it as neatly as you guys do? It looks so much better then my typing =).
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  5. #5
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    Quote Originally Posted by ninjahcamp3r View Post
    Alright the last part is also what I have but is it all the way simplified.

    I also forgot to include that a_n can also be expressed as (n + 1)! - n!.

    Could you also give me a link on how you guys right it as neatly as you guys do? It looks so much better then my typing =).
    These formulas are written using LaTex coding. There is a section in here that may help:
    http://www.mathhelpforum.com/math-help/latex-help/
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