A summation series of binomial coefficients

Question

Given.

$$(1 + x^{2005} + x^{2006} + x^{2007})^{2008} = A_0 +A_1x +A_2x^2 + \cdots + A_nx^n$$

We are required to calculate $A_0 - A_1/3-A_2/3 + A_3 -A_4/4 -A_5/5 + A_6 - \cdots$

I tried approaching the problem by setting $x$ to 1 and $ω$ in two different cases.

For $x=1$, we get sum of all coefficients as $4^{2008} $

For $x=ω$ , we get :$- (A_0 + A_3 +A_6 + \cdots) -\frac12(A_1+A_2+A_4+A_5+\cdots) =1$

(Since setting $x = ω$ in the given equation up top, we get LHS${}={}$RHS${}=1$, thus comparing the real parts we get the result)

But solving this two equations simultaneously gives me a wrong result. Would appreciate if anyone could point out the mistake in my approach.

Answer 1

Here is one way to represent the coefficients $A_k$ of the polynomial $A(x)$ \begin{align*} A(x)=(1 + x^{2005} + x^{2006} + x^{2007})^{2008} = \sum_{j=0}^{2007\cdot2008}A_jx^j \end{align*}

We obtain \begin{align*} A(x)&=(1 +x^{2005} + x^{2006} + x^{2007})^{2008}\\ &=\sum_{j=0}^{2008}\binom{2008}{j}(x^{2005}+x^{2006}+x^{2007})^j\\ &=\sum_{j=0}^{2008}\binom{2008}{j}x^{2005j}(1+x+x^2)^j\\ &=\sum_{j=0}^{2008}\binom{2008}{j}x^{2005j}\sum_{{i_1+i_2+i_3=j}\atop{i_1,i_2,i_3\geq 0}}\binom{j}{i1,i2,i3}1^{i_1}x^{i_2}x^{2i_3}\tag{1}\\ &=\sum_{j=0}^{2008}\binom{2008}{j}\sum_{{i_1+i_2+i_3=j}\atop{i_1,i_2,i_3\geq 0}}\frac{j!}{i_1!i_2!i_3!}x^{2005j+i_2+2i_3}\\ \end{align*}

Comment:

• In (1) we use the expansion with trinomial coefficients.

Note, the polynomial $A(x)$ is sparse. If we write $A(x)$ in the form \begin{align*} A(x)&=(1 + x^{2005} + x^{2006} + x^{2007})^{2008}\\ &=\binom{2008}{0}+\binom{2008}{1}x^{2005}(1+x+x^2)\\ &\qquad\qquad\quad+\binom{2008}{2}x^{2005\cdot2}(1+x+x^2)^2\\ &\qquad\qquad\quad+\cdots\\ &\qquad\qquad\quad+\binom{2008}{2007}x^{2005\cdot 2007}(1+x+x^2)^{2007}\\ &\qquad\qquad\quad+x^{2005\cdot2008}(1+x+x^2)^{2008} \end{align*}

we see that the expansion of $A(x)$ starts with

\begin{align*} A(x)=1+2008x^{2005}+2008x^{2006}+2008x^{2007}+\frac{2008\cdot 2007}{2}x^{4010}+\cdots \end{align*} so, \begin{align*} &A_1=A_2=\ldots=A_{2004}=0\\ &A_{2008}=A_{2009}=\ldots A_{4009}=0 \end{align*}

Note: A pattern of the coefficients in OPs expression \begin{align*} A_0-\frac{1}{3}A_1-\frac{1}{3}A_2+A_3-\frac{1}{4}A_4-\frac{1}{5}A_5+A_6-\ldots \end{align*} is not obvious.