Simplify the sum $$ \sum_{ \substack{ i, j, k \geq 0 \text{ and } i+j+2k=r }} (-1)^i \binom{n}{i} \binom{n}{j} \binom{n}{k} $$ for n, r $\in$ N .
I know that it can be simplified to $$ [x^r] (1+x)^n(1-x)^n(1+x^2)^n = [x^r] (1-x^2)^n(1+x^2)^n = [x^r] (1-x^4)^n $$ which gives $\binom{n}{\frac{r}{4}} (-1)^{\frac{r}{4}}$ if $4|r$ and $0$ otherwise.
But how do I get to that point?
I tried writing $$ \sum_{n} \left( \sum_{i} (-1)^i \binom{n}{i} \sum_{j} \binom{n}{j} \sum_{k} \binom{n}{k} \right)x^n $$
so I can use binomial theorem, but then I would need three $x^n$'s total to use theorem three times.
Thanks.
$$\begin{align}(1+x)^n(1-x)^n(1+x^2)^n&=\sum_{j=0}^n\binom njx^j \sum_{i=0}^n\binom ni(-x)^i \sum_{k=0}^n\binom nk(x^2)^k\\&= \sum_{r=0}^{4n}x^r\sum_{i, j, k \ge0\atop i+j+2k=r}(-1)^i\binom ni\binom nj\binom nk \end{align}$$