Find the coefficient of $x^{12}$ in $(1+x)^{-1}$

Question

My work:

$(1 + x)^{-1} = \left(\cfrac{1-x^2}{1-x}\right)^{-1} = \left(\cfrac{1-x}{1-x^2}\right)^{1} =(1-x)^1(1-x^2)^{-1}$

So we do $C(1,0) \cdot C(-1,6) = 0$ to find all the ways to get $x^{12}$

But the final answer is $1$ not $0$.

What am I doing wrong?

Answer 1

Remember that $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots$$ for all $\left| t\right| <1$. Substitute $t=-x$.