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

Se foloseste formula 1+2+3+..+n= n*(n+1)/2
Astfel :
a) S=20*(20+1)/2=10*21=210  
b) S=49*(49+1)/2=49*25=1225
c) S=100*(100+1)/2=50*101=5050
d) S=200*(200+1)/2=100*201=20100

Answer 2

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