Finding $ \sum_{k=0}^n (-1)^k A_k {n \choose k}, ~\text{if}~ (1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$

Question

Finding $S=\sum_{k=0}^n (-1)^k A_k{n\choose k}$, if $(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k.$ \begin{align} (1+x+x^2)^n &= A_0+A_1x+A_2x^2+\dots+A_nx^n+\dots+A_{2n}x^{2n} \\ (1-1/x)^n &= {n \choose 0}-{n \choose 1}x^{-1}+{n\choose 2}x^{-2}+\dots+(-1)^nx^{-n} \end{align}

Multiplying these two we get $S=\sum_{k=0}^{n} (-1)^k A_k {n \choose k}=\text{Coefficient of $x^0$ in } (x^3-1)^n/x^n$

Or $S=\text{Coefficient of $x^n$ in } (x^3-1)^n=0$, when $n$ is not a multiple of $3$.

The question is how else we can prove this result? How to modify this result when $n$ is a multiple of $3$ (that is, $n=3p$)?

EDIT I find that if $n=3p$, then $S= \text{Coefficient of}~ x^{3p}$ in $(x^3-1)^{3p}={3p \choose p}$

Answer 1

This is not a different approach, but a convenient notation which might be useful. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n(-1)^kA_k\binom{n}{k}} &=\sum_{k=0}^n\binom{n}{k}(-1)^k[x^k]\left(1+x+x^2\right)^n\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k[x^k]\left(\frac{x^3-1}{x-1}\right)^n\\ &=[x^0]\left(\frac{x^3-1}{x-1}\right)^n\sum_{k=0}^n\binom{n}{k}\left(-\frac{1}{x}\right)^k\tag{1}\\ &=[x^0]\left(\frac{x^3-1}{x-1}\right)^n\left(1-\frac{1}{x}\right)^n\tag{2}\\ &=[x^n]\left(x^3-1\right)^n\tag{3}\\ &=[x^n]\sum_{q=0}^n\binom{n}{q}x^{3q}(-1)^{n-q}\\ &\,\,\color{blue}{= \begin{cases} \binom{n}{n/3}&\quad 3\ |\ n\\ 0&\quad 3\not|\ n \end{cases}}\tag{4} \end{align*}

Comment:

• In (1) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (2) we apply the binomial theorem.

• In (3) we simplify the expression and apply the same rule as in (1) again. We expand the binomial in the following line.

• In (4) we select the coefficient of $x^n$.