Solve the equation, hat consists on an arithmetical progression.

Question

$$1+x+x^2+x^3+\cdots+x^{99}=0.$$

I said to prove with $0+1+2+3+\cdots+99=0$. How should I proceed?

Answer 1

$1+x+\cdots+x^{99}=\frac{x^{100}-1}{x-1}=0\iff \begin{cases}x^{100}=1\\x\neq 1\end{cases}\iff x=-1$ (if $x\in\mathbb R$).

This uses the well-known geometric progression formula: $$a+ar+ar^2+\cdots+ar^{m-1}=\frac{a(r^m-1)}{r-1}$$

Answer 2

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 ! $$1+x^2+x^4+...+x^{98} \neq 0\\$$