$$\frac 1{1+x+x^2}$$ $$ = \sum^\infty_{n=0} {(-1)}^n{(x+x^2)}^n$$ $$ = \sum^\infty_{n=0}{(-x)}^n \sum^n_{k=0} {_nC_k}x^k$$ $$ = \sum^\infty_{n=0}\sum^{[\frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$$
I wanted to simplify the last expression to an expression with a single $\sum$.
So I tried substituting a few terms.
$$\begin{align} \tag{n=6} \sum^3_{k=0}{(-1)}^{6-k}{_{6-k}C_k} = 1 \\ \tag{n=7} \sum^3_{k=0}{(-1)}^{7-k}{_{7-k}C_k} = -1 \\ \tag{n=8} \sum^4_{k=0}{(-1)}^{8-k}{_{8-k}C_k} = 0 \\ \tag{n=9} \sum^4_{k=0}{(-1)}^{9-k}{_{9-k}C_k} = 1 \\ \end{align} $$
Then, is there any generic simplified form of $\sum^{[\frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$?
Hint for a simpler approach. Note that for $|x|<1$, $$\frac{1}{1+x+x^2}=\frac{1-x}{1-x^3}=(1-x)\sum_{k=0}^{\infty} x^{3k}.$$ Can you take it from here?