Question

1+2+3+...+1199+1200=

Answer 1

1+2+3+....+1199+1200
⇒1200·1201/2
⇒600·1201
⇒720600;

Answer 2

$1+2+3+....+1199+1200= \frac{1200(1200+1)}{2} =600*1201=720600$

$\text{S-a folosit formula:}\\ \\ \boxed{\frac{n(n+1)}{2} } \\ unde \ n \ este \ ultimul \ numar \ din \ sir$