find the sum to $n$ terms of the series $1+4w+9w^2+...+n^2w^{n-1}$ where $w$ is $n$th root of unity I want to find the sum to $n$ terms of the series
$$1+4w+9w^2+...+n^2w^{n-1}$$
where$w$ is $n$th root of unity.
Let $$S_n = 1+4w+9w^2+...+n^2w^{n-1}$$
then $$ wS_n=w+4w^2+....+(n-1)^2w^{n-1}+n^2$$
therefore $$(1-w)S_n=1+3w+5w^2+...+(2n-1)w^{n-1}-n^2$$
Now when I find the sum of $$ 1+3w+5w^2+...+(2n-1)w^{n-1}$$ it comes out to be $$\frac{-2n}{1-w}$$ and after substituting we get $$ S_n= \frac{-2n}{(1-w)^2}-\frac{n^2}{1-w}$$
The problem is my book doesn't agree with my answer 
The answer is $$\frac{n[(1-w)n+2]}{3w}.$$

I find  $(1-w)S_n=-n^2+1+3w+5w^2+...+(2n-1)w^{n-1}$
Let $U_n=1+3w+5w^2+...+(2n-1)w^{n-1}$
$(1-w)U_n=1+2(w+w^2++w^{n-1})-(2n-1)=-2n+2(1+w+w^2++w^{n-1})$
Now $1+w+w^2++w^{n-1}=\dfrac{1-w^n}{1-w}=0$
