Solve the equation, hat consists on an arithmetical progression. $$1+x+x^2+x^3+\cdots+x^{99}=0.$$
I said to prove with $0+1+2+3+\cdots+99=0$. How should I proceed?

another observation $$1+x+x^2+x^3+...+x^{99}=\\(1+x^2+x^4+...+x^{98})+(x+x^3+x^5+...+x^{99})=0\\(1+x^2+x^4+...+x^{98})+x(1+x^2+x^4+...+x^{98})=\\(1+x^2+x^4+...+x^{98})(1+x)=0\\\rightarrow (1+x)=0 \rightarrow x=-1$$note that ! $$1+x^2+x^4+...+x^{98} \neq 0\\$$
