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

Question

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 $$\frac{n[(1-w)n+2]}{3w}.$$

With $n=1$ your book answer doesn't agree with $S_1$. – Nosrati Mar 12 '17 at 13:33 — Nosrati, Mar 12 '17 at 13:33
I agree with your answer. I think the book must be wrong. – TonyK Mar 12 '17 at 14:16 — TonyK, Mar 12 '17 at 14:16
The book answer also doesn't agree with $S_2=1+4(-1)=-3$. – Barry Cipra Mar 12 '17 at 14:24 — Barry Cipra, Mar 12 '17 at 14:24

score 2 · Answer 1 · answered Mar 12 '17 at 14:00

2

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 $1+w+w^2++w^{n-1}=\dfrac{1-w^n}{1-w}=0$

answered Mar 12 '17 at 14:00

lab bhattacharjee

274,582

I edited my question. thank you sir. But still the root problem is there. – Rayees Ahmad Mar 12 '17 at 14:11
1

@RayeesAhmad, Welcome! I think, we can safely focus on the correctness of the approach, the correct answer should follow. – lab bhattacharjee Mar 12 '17 at 14:12
Yes, that is undoubtedly true!!! – Rayees Ahmad Mar 12 '17 at 14:15

student · Answer 2 · 2017-04-16T13:28:14.797

As you showed above, $$(1-w)S_{n}=1+3w+5w^2+\cdots +(2n-1)w^{n-1}-n^2.$$ Thus, $$(1-w)S_n +n^2 = 1+3w+5w^2+\cdots +(2n-1)w^{n-1}.$$ We now need the following fact (which you allude to, and I can prove but it might make this answer very long) $$ 1+3w+5w^2+\cdots +(2n-1)w^{n-1} = \frac{2n}{w-1}.$$ Combining these identities, we get $$(1-w)S_n +n^2 = \frac{2n}{w-1},$$ so $$(w-1)S_n - n^2 = -\frac{2n}{w-1},$$ and solving for $S_n$, we get $$S_n = \frac{n^2}{(w-1)} - \frac{2n}{(w-1)^2}.$$ The answer in the above form appears as an exercise in the book Complex Function Theory by Donald Sarason (page 8), and so I am sure it is correct.

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

2 Answers2