1+4+7+...+301 = 1+1+3+1+3×2+1+3×3+...+1+3×100 = 1+1+...+1+3(1+2+3+...+100)
1+1+...+1 sunt 101 de numere
[tex] \frac{101+3(1+2+3+...+100)}{1+2+3+...+100} = \frac{151}{50} [/tex]
[tex] \frac{101+3}{1} = \frac{151}{50}[/tex]
[tex]104 = \frac{151}{50} [/tex]
Am gresit cumva undeva?