Question

Sa se rezolve ecuatiile:
a) 1+x+x²+x³+...+x⁹⁹=0
b) 1+x+x²+x³+...x¹⁰⁰=0

Answer 1

$1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}$

$1+x+x^2+...+x^{99} = 0$
$\frac{1-x^{100}}{1-x} = 0$
CE: x =/= 1
$1-x^{100} = 0$
$x^{100} = 1$
$x^n = f(x)$, n este numar par => $x = \sqrt[n]{f(x)} ; -\sqrt[n]{f(x)}$
$x_1 = 1^{ \frac{1}{100}};x_2=-1^{ \frac{1}{100} }$
$x_1 = 1$ - Nu convine
$x_2 = -1$