Question

stabiliti daca 10 divide pe a , unde a = 1x2+2x3+3x4+4x5+....... +2010x2011

Answer 1

A=1x2+2x3+3x4+...+2010x2011
u(A)=u[u(1x2)+u(2x3)+...+u(2010x2011)
u(A)=u[2+6+2+0+0+2+6+...+0+0]
Număr termeni : (2010-1)/1+1=2010
Grupăm in termeni de câte 5 “(2+6+2+0+0)”= 2010:5=402 grupe
u(A)=u[u(402(2+6+2+0+0))]
u(A)=u[u(402x10)]
u(A)=u(4020)
u(A)=0
Dar 10|A <=> u(A)=0 => 10|A



Sper ca te-am ajutat ! :)