$$ (x^{2}+x+1)^{100}=a_0+a_1x+a_2x^{2}+...+a_{199}x^{199}+a_{200}x^{200}$$ $$\sum_{i=1}^{200} {1\over{1+x_i}} =?$$ Can somebody help me? Thank you!


closed as off-topic by Servaes, Parcly Taxel, Lord Shark the Unknown, user8795, Zain Patel May 14 '17 at 14:49

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  • $\begingroup$ Do you mean $a_i$ instead of $x_i$ in the sum? $\endgroup$ – samjoe May 14 '17 at 12:15
  • $\begingroup$ No. It's $x_{i}$ in the sum. $\endgroup$ – Andrew May 14 '17 at 12:18
  • $\begingroup$ That doesn't make sense. You haven't defined the $x_i$. Also, what are your thoughts on the problem? $\endgroup$ – Servaes May 14 '17 at 12:19
  • $\begingroup$ You forgot to mention what you have tried. $\endgroup$ – StubbornAtom May 14 '17 at 12:19
  • $\begingroup$ $x_{i}$ are the solution of the problem, $x_1 ,x_2...x_{200}$ . I tried to amplified each fraction for a common denominator. But I don't know what i can do to solve it after that $\endgroup$ – Andrew May 14 '17 at 12:21

$$\sum_{k=1}^{200}\frac{1}{1+x_k}=100\left(\frac{1}{1+x_1}+\frac{1}{1+x_2}\right)=\frac{100(2+x_1+x_2)}{1+x_1+x_2+x_1x_2}=$$ $$=\frac{100(2-1)}{1-1+1}=100.$$

  • $\begingroup$ Can you explain me why is 100 * paranthesis? $\endgroup$ – Andrew May 14 '17 at 12:33
  • $\begingroup$ @Andrew Because our polynomial has $100$ roots $x_1$ and $100$ roots $x_2$, where $x_1$ and $x_2$ are roots of the polynomial $x^2+x+1$. You are welcome! $\endgroup$ – Michael Rozenberg May 14 '17 at 12:45

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