Question

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

rezolvare completa va rog

Answer 1

1+2+3+.....+1199+1200=
=1200*(1200+1):2=
=1200*1201:2=
=600*1201=720600

Answer 2

$\displaystyle \boxed{ \frac{n(n+1)}{2} } \\\\ \text{n=ultimul numar din sir } \\\\ 1+2+3+.....+ 1199+1200= \frac{1200(1200+1)}{2}= \\ \\ \\ \frac{\not1200\cdot1201}{\not2}= 600 \cdot 1201 \\ \\ \\ 600\cdot1201=720600$