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A high school polynomial root question:

Let $W(x)=x^{2n} + x^{2n-1} + x^{2n-2} + \cdots + x + 1$. Let the root of $W(x)$ be $r_1, r_2, ... r_{2n}$. Find $$ \sum_{i=1}^{2n}\frac{r_i^2}{r_i+1} $$ in terms of $n$.

What I found:

  • $r_i^{2n+1}=1$ according to the sum of geometry sequence
  • Feel it has some relation with Vieta's formula, but don't know how to use it
  • It's easily to see $$ 1 + \frac{r_i^2}{1+r_i}+r_i^3+r_i^5+...+r_i^{2n-1}=0 $$ But I can't see how the $r_i^3, r_i^5...r_i^{2n-1}$ factored out by $n$
xwa130
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  • Welcome to Math.SE! ... The community prefers/expects a question to include something of what the asker knows about the problem. (What have you tried? Where did you get stuck? etc) This helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already understand or using techniques beyond your skill level. (It also helps convince people that you aren't simply trying to get them to do your homework for you. An isolated problem statement with no evidence of personal effort makes a poor impression, attracting down- and close-votes.) – Blue Feb 03 '24 at 18:41
  • Thanks @Blue, updated the question with my findings. – xwa130 Feb 03 '24 at 18:54

2 Answers2

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If $r_i$ is a root of $W$ then $1/r_i$ is also a root of $W$, and these two roots are different (because $W(\pm 1) \ne 0$). Therefore the set of $2n$ roots can be written as $$ r_1, r_2, \ldots, r_n, \frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_n} $$ and then $$ \sum_{i=1}^{2n}\frac{r_i^2}{r_i+1} = \sum_{i=1}^{n}\left(\frac{r_i^2}{r_i+1} +\frac{1/r_i^2}{1/r_i+1} \right) \\ = \sum_{i=1}^{n}\left( r_i + \frac{1}{r_i} - 1 \right) = -n + \sum_{i=1}^{2n} r_i = -n -1 \, . $$

Martin R
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First note that $$\frac{x^2}{x+1} = x-1+\frac{1}{1+x}$$ So that your sum becomes

$$S = -1 - 2n + \sum_{k=1}^{2n} \frac{1}{r_k+1}$$

Now $$\sum_{k=0}^{2n} \frac{1}{r_k+1} = \sum_{k=1}^n \left(\frac{1}{r_k+1} + \frac{1}{r_{2n+1-k}+1}\right) = n$$

Since $r_j\cdot r_{2n+1-j}=1$, and there's no value where $j=2n+1-j$.

Therefore $$\boxed{S=-1-2n+n=-1-n}$$

Sahaj
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