Question

. Calculati numerele : S=1+2+3+...+10+11

2) S=1+2+3+...+50+51

3) S=1+2+3+...+89​




DAU COROANA

Answer 1

La toate folosești formula lui Gauss

$1 + 2 + 3 + ... + x = \\ \\ \huge= \frac{x \times (x + 1)}{2}$

unde x este ultimul număr

1)

$1 + 2 + 3 + ... + 10 + 11 = \\ = \frac{11 \times 12}{2} = 11 \times 6 = 66$

2)

$1 + 2 + 3 + .. + 50 + 51 = \\ = \frac{51 \times 52}{2} = 51 \times 26 = 1326$

3)

$1 + 2 + 3 + ... + 89 = \\ = \frac{89 \times 90}{2} = 89 \times 45 = 4005$

Answer 2

Răspuns:

2 S=1+2+3....+50+51 = 50 x(50+1) :2 = 50 x 51 :2 =2550 :2 =1275

3 S=1+2+3......+89= 89 ×(89+1):2= 89×90 :2 =8010 :2 = 4005

Am folosit suma lui Gauss iar la aceasta se pune

1+2+3....+N si se face N×(N+1):2