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Math Help - Challenging multinomial theorem question?

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


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

    We have: <br />
\sum\limits_{k = 0}^{s - 1} {F\left( {e^{\tfrac{{2 \cdot \pi }}<br />
{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 }}<br />
{s} \cdot i \cdot k} } \right)^n } } \right] \cdot x^n } <br />

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

    Now set s=4 and F(x)=(1+x+x^2+x^3)^{4n} to get your answer, by the way, if x\ne 1 you can write <br />
F\left( x \right) = \left( {\tfrac{{1 - x^4 }}<br />
{{1 - x}}} \right)^{4n} <br />
which is very suggestive since we are working with 4th roots of unity
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