Trinomial expansion variation - generalize? One may represent $(1+x+x^2)^k$ = $\sum_{\ell=0}^{2k}\begin{pmatrix}k\\\l-k\end{pmatrix}_2x^\ell$, where $()_2$ is the trinomial coefficient. Any one with experience how to represent $(1+x+0.5x^2)^k$ in a similar manner? Thanks.

${ \bf Edit}$: A little searching allowed me to find $$ (a+b+c)^n = \sum_{i+j+k=n} \frac{ n!}{i!j!k!} a^i b^j c^k $$ Thus $$ ( x^2/2 + x +1)^n = \sum_{i+j+k=n} \frac{ n!}{2^i i!j!k!} x^{2i+j}$$

${ \bf Original Post}$: You can use the regular binomial theorem to obtain this. Factor the quadratic into it's roots (they may be complex valued). Then apply the binomial theorem: i.e. $$ (ax^2 + bx + c )^n = a^n( x-\lambda_+ )^n ( x - \lambda_-)^n $$ where $$ \lambda_\pm = \frac{ -b \pm \sqrt{ b^2 -4ac}}{2a} $$ In your example, we have $$ ( x^2/2 + x + 1)^n= \frac{1}{2^n}( x - (-1 +i) )^n( x - ( -1 -i))^n$$ Then apply the binomial theorem, and use the following rule to multiply the series together: $$ \sum_{k} a_k \sum_l b_l = \sum_{n} \left ( \sum_{j=0}^n a_j b_{n-j} \right) $$