Question

2 plus 4 plus 6 plus 80 împărțit la 1 plus 2 plus 3 plus 40 egal cu​

Answer 1

 

$\displaystyle\bf\\(2+4+6+...+80) : (1+2+3+...+40)=\\\\=\frac{(2+4+6+...+80)}{(1+2+3+...+40)}=\frac{2(1+2+3+...+40)}{(1+2+3+...+40)}=\boxed{\bf2}\\\\Explicatii:\\Primul~sir~contine~doar~numere~pare.\\Dupa~ce~am~scris~impartirea~ca~o~fractie,\\~am~dat~pe~2~factor~comun~la~numarator~si~am~ramas~in~paranteza\\cu~un~sir~identic~cu~cel~de~la~numitor.\\Am~simplificat~si~am~ramas~cu~2$