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Consider the polynomial $P(x) = (x^2 + x + 1)^{2015} + x + 1$ with the roots $x_k, \: 1 \le k \le 4030$.

Evaluate

$$\sum _{k = 1}^{4030} \frac{1}{x_k}$$

I have found that $P(i) = 1$ and the remainder of $P / (x^2 + 1)$ is also $1$.

I think the sum should be $$\frac{-a_1}{a_0}$$ where $a_0$ is the coefficient of $x^{0}$, which is $2$, but I don't know how to find the other coefficient( that of $x^1$ )

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  • $\begingroup$ You're right. How many ways do you get an $x^1$ when you multiply out the power of $(x^2+x+1)$? It seems to me that you might just ignore the $x^2$? $\endgroup$ Apr 26, 2017 at 9:59
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    $\begingroup$ $P'(x) = 2015(x^2+x+1)^{2014}(2x+1) + 1$ and $P'(0) = 2016$. Thus coefficient of $x$ is 2016. $\endgroup$
    – user348749
    Apr 26, 2017 at 10:25
  • $\begingroup$ @Muralidharan You can post as answer, thank you ! $\endgroup$
    – Liviu
    Apr 26, 2017 at 10:34

1 Answer 1

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The easiest way to find $a_1$, the coefficient of $x$ is to use differentiation. We have $$P'(x) = 2015(x^2+x+1)^{2014}(2x+1) + 1$$ and putting $x=0$, we get $P'(0) = a_1 = 2016$

One can also expand as follows: Writing $P(x) = ((x^2+1)+x)^{2015} + x+1$ and expanding the first term using Binomial theorem, the only term that contains $x$ in the expansion is $\binom{2015}{2014} x (x^2+1)^{2014}$ and hence the coefficient is $2015+1=2016$

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