Question

va rog ajutor 2003+2.(1+2+...+2002)

Answer 1

2003+2x(2003x2002/2)= 2003+2x2005003= 2003+401006=403009

Answer 2

2003+2*(1+2+...+2002)=
2003+2* 2005003=
2003+4010006=
4012009

1+2+...+2002=$\frac{n*(n+1)}{2}$
1+2+...+2002=$\frac{2002*(2002+1)}{2}$ ⇒ $\frac{4010006}{2} = 2005003$