Question

Rezolvaţi în Q existenţă ecuaţia:
x+2x+3x+...+100x=-1+2-3+4-...-201+202 .

Answer 1

$x(1+2+3+\ldots +100)=(-1+2)+(-3+4)+\ldots +(-201+202)$
$x\cdot\displaystyle\frac{(1+100)\cdot 100}{2}=\underbrace{1+1+\ldots +1}_{\mbox{101 ori}}$
$50\cdot 101x=101\Rightarrow 50x=1\Rightarrow x=\displaystyle\frac{1}{50}\in\mathbb{Q}$