Question

Calculati sumele
a) S=1+2+3+....+20

b) S=1+2+3+...+49

c) S=1+2+3+....+100

d) S=1+2+3+....+200

VA ROG AJUTATIMA DAU COROANA VA ROOOOGG

Answer 1

a)S=1+2+3+....+20=20*21/2=10*21=210; B)S= 1+2+3+...+49=49*50/2=49*25=1225 C)S= 1+2+3+4+....+100=100*101/2=50*101=5050 D) S=1+2+3+....+200=200*201/2=100*201=20100 - la toate am folosit formula 1+2+3+...+n=[n*(n+1)]/2

Answer 2

$\displaystyle a).1+2+3+...+20= \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = \frac{420}{2} =210 \\ \\ b).1+2+3+...+49= \frac{49(49+1)}{2}= \frac{49 \times 50}{2} = \frac{2450}{2} = 1225 \\ \\ c).1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} =5050 \\ \\ d).1+2+3+...+200= \frac{200(200+1)}{2} = \frac{200 \times 201}{2} = \frac{40200}{2} =20100$