Question

1+2+3...200
1+2+3..250
4+8+12...200
5+10+15...600

Answer 1

$\displaystyle a).1+2+3+...+200= \frac{200(200+1)}{2} = \frac{200 \times 201}{2} = \frac{40200}{2} =20100 \\ \\ b).1+2+3+...+250= \frac{250(250+1)}{2} = \frac{250 \times 251}{2} = \frac{62750}{2} =31375 \\ \\ c).4+8+12+...+200=4(1+2+3+...+50)= \\ \\ =4 \times \frac{50(50+1)}{2} =4 \times \frac{50 \times 51}{2} =4 \times \frac{2550}{2} =4 \times 1275=5100$

$\displaystyle d).5+10+15+...+600=5(1+2+3+...+120)= \\ \\ =5 \times \frac{120(120+1)}{2} =5 \times \frac{120 \times 121}{2} =5 \times \frac{14520}{2} =5 \times 7260=36300$

Answer 2

Se aplica Formula lui Gauss:
1+2+3+...+n=n(n+1)/2⇒
1+2+3+...+200=200·(200+1)/2=20100
1+2+3+...+250=250(250+1)/2 =31375
4+8+12+...+200=4(1+2+3+...50)=4.50·(50+1)/2=5100
5+10+15+...+600=5(1+2+3+...+120)=5·120(120+1)/2=36300