# 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 $\displaystyle F\left( x \right) = \sum\limits_{n = 0}^{ + \infty } {a_n \cdot x^n } {\text{ and }}s \in \mathbb{Z}^ +$ 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