# Thread: Challenging multinomial theorem question?

1. ## 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.)

How to solve this?

2. Proof
Given $
F\left( x \right) = \sum\limits_{n = 0}^{ + \infty } {a_n \cdot x^n } {\text{ and }}s \in \mathbb{Z}^ +
$
we have: $
\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: $
\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 $
m = \dot s
$
we have $
\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 $
a \cdot s = m
$
thus $\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 $
m \ne \dot s
$
we can consider the geometric sum: $
\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: $
\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: $
\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 $s=4$ and $F(x)=(1+x+x^2+x^3)^{4n}$ to get your answer, by the way, if $x\ne 1$ you can write $
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