<s>[INST] Polynomial sum relation Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)
 [/INST] 
Is there a general solution for this type of problem

Consider for example $P(x)=(x^2+x+1)^n\,$.

$\sum_{k=1}^{2n} \frac{1}{x_k}$

Let $y_k = \frac{1}{x_k} \ne 0\,$ then $P(\frac{1}{y_k})=P(x_k)=0\,$, so $y_k$ are the roots of: $$Q(y)=y^{2n} P\left(\frac{1}{y}\right)=y^{2n}\left(\frac{1}{y^2}+\frac{1}{y}+1\right)^n=(1+y+y^2)^n=y^{2n}+\binom{n}{0,1,n-1}y^{2n-1}+\cdots\,$$
Then by Vieta's relations $\sum_{k=1}^{2n} y_k=-\binom{n}{0,1,n-1}=-n\,$.

$\sum_{k=1}^{2n} \frac{1}{1+x_k}$

Let $z_k=x_k+1\,$, then $P(z_k-1)=P(x_k)=0\,$, so $z_k$ are the roots of:
$$\require{cancel}
R(z) = P(z-1)=\left((z-1)^2+(z-\bcancel{1})+\bcancel{1}\right)^n=(z^2-z+1)^n
$$
Then, by an argument entirely similar to the above, it follows that $\sum_{k=1}^{2n} \frac{1}{z_k}=\binom{n}{0,1,n-1}=n\,$.

find the coefficient of $a_1$ and remainder for $P(x)/(x^2+x)$

$$
P(x)=\big((x^2+x)+1\big)^n=(x^2+x)^n+\binom{n}{1}(x^2+x)^{n-1}+\cdots+\binom{n}{n-1}(x^2+x) + 1
$$
All terms are multiples of $x^2+x$ except the very last one, so the remainder of the division is $1\,$ which is a constant polynomial so $a_1=0\,$.
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