Question

la 8 ca rogggggggggggg

Answer 1

$\frac{4 + 8 + 12 + ... + 204}{3 + 6 + 9 + ... + 153} \\ \\ = \frac{4(1 + 2 + 3 + ... + 51)}{3(1 + 2 + 3 + ... +51) } \\ \\ \text{sumele \: 1 + 2 + ... + 51 se simplifica} \\ \text{deci ne ramane} \: \frac{4}{3}$