Results 1 to 2 of 2

Thread: Challenging multinomial theorem question?

  1. #1
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    Challenging multinomial theorem question?

    If (1 + x + x^2 + x^3)^4n = ∑( r = 0 to 12n) ar. x^r, then find the value of ∑(r = 0 to 3n) a(4r).

    (NOTE: In ar, r is a subscript.
    In a(4r), 4r is a subscript.)


    Answer: 2^(8n - 2)

    How to solve this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Proof
    Given $\displaystyle
    F\left( x \right) = \sum\limits_{n = 0}^{ + \infty } {a_n \cdot x^n } {\text{ and }}s \in \mathbb{Z}^ +
    $ we have: $\displaystyle
    \boxed{\tfrac{1}
    {s} \cdot \sum\limits_{k = 0}^{s - 1} {F\left( {e^{\tfrac{{2 \cdot \pi }}
    {s} \cdot i \cdot k} \cdot x} \right)} = \sum\limits_{\left. s \right|n} {a_n \cdot x^n } }
    $
    .......
    We will show that: $\displaystyle
    \sum\limits_{k = 0}^{s - 1} {\left( {e^{\tfrac{{2 \cdot \pi }}
    {s} \cdot i \cdot k} } \right)^m } = \left\{ \begin{gathered}
    0{\text{ if }}m \ne \dot s{\text{ }} \hfill \\
    s{\text{ if }}m = \dot s \hfill \\
    \end{gathered} \right.,m \in \mathbb{N}
    $


    • If $\displaystyle
      m = \dot s
      $ we have $\displaystyle
      \left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot k} } \right)^m = e^{a \cdot 2 \cdot \pi \cdot i \cdot k} ,a \in \mathbb{N}
      $ where $\displaystyle
      a \cdot s = m
      $ thus $\displaystyle \left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot k} } \right)^m = e^{a \cdot 2 \cdot \pi \cdot i \cdot k} = 1 \Rightarrow \sum\limits_{k = 0}^{s - 1} {\left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot k} } \right)^m } = s
      \$
    • Otherwise, if $\displaystyle
      m \ne \dot s
      $ we can consider the geometric sum: $\displaystyle
      \sum\limits_{k = 0}^{s - 1} {\left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot k} } \right)^m } = \sum\limits_{k = 0}^{s - 1} {\left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot m} } \right)^k } = \frac{{\left( {e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot m} } \right)^s - 1}}
      {{e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot m} - 1}} = \frac{{e^{2 \cdot \pi \cdot i \cdot m} - 1}}
      {{e^{\tfrac{{2 \cdot \pi }}
      {s} \cdot i \cdot m} - 1}} = 0
      $

    We have: $\displaystyle
    \sum\limits_{k = 0}^{s - 1} {F\left( {e^{\tfrac{{2 \cdot \pi }}
    {s} \cdot i \cdot k} \cdot x} \right)} = \sum\limits_{n = 0}^{ + \infty } {\left[ {a_n \cdot \sum\nolimits_{k = 0}^{s - 1} {\left( {e^{\tfrac{{2 \cdot \pi }}
    {s} \cdot i \cdot k} } \right)^n } } \right] \cdot x^n }
    $

    Thus, by what we've just proven: $\displaystyle
    \sum\limits_{k = 0}^{s - 1} {F\left( {e^{\tfrac{{2 \cdot \pi }}
    {s} \cdot i \cdot k} \cdot x} \right)} = s \cdot \sum\limits_{\left. s \right|n} {a_n \cdot x^n } \blacksquare
    $

    Now set $\displaystyle s=4$ and $\displaystyle F(x)=(1+x+x^2+x^3)^{4n}$ to get your answer, by the way, if $\displaystyle x\ne 1$ you can write $\displaystyle
    F\left( x \right) = \left( {\tfrac{{1 - x^4 }}
    {{1 - x}}} \right)^{4n}
    $ which is very suggestive since we are working with 4th roots of unity
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multinomial question....
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: Jan 22nd 2011, 05:00 AM
  2. Multinomial coefficient question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Aug 29th 2009, 08:21 PM
  3. A Challenging Question
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: Aug 15th 2009, 06:03 PM
  4. Inductive Proof: Multinomial Theorem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Sep 3rd 2008, 06:37 PM
  5. Challenging--- Central Limit Theorem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Apr 22nd 2008, 07:12 PM

Search Tags


/mathhelpforum @mathhelpforum