Question

Rezolvați 3(1+2+3+...+1000)+(3+4+5+...+1000)*(-3)

Answer 1

Prin regula lui Gauss avem: $\frac{n(n+1)}{2}$
3($\frac{1000*1001}{2}$)+($\frac{997*998}{2}$)*(-3)
3*(550*1001)+(499*997)*(-3)=3*550550+497503*(-3)=1651650-1492509=159141

(3+4+5+...+1000) l-am scris ca (1+2+3+...+997).