Question

=125+126+127+......+2000)-(1+5+5+.....+65) va rog ajutațimă dau 25 puncte

Answer 1

1+2+3+...+2000=2000x2001:2=2001000
1+2+3+...+124=124x125:2=7750
125+126+127+...+2000=2001000-7750=1993250

(1+5+5+.....+65) ?? nu stiu

Answer 2

S=125+126+127+......+2000=

varianta I - prin diferente de sume Gauss:

S1=1+2+3+...+2000=(2000×2001):2=2001000

S2=1+2+3+...........+124=(124×125):2=7750

S=S1-S2=2001000-7750=1993250

varianta a II a suma reprezinta o progresie aritmetica

125+126+127+......+2000 progresie aritmetica, a1=125, an=2000, r=1

an=a1+(n-1)×r, 2000=125+(n-1)⇒2000-125+1=n;  n=1876

S=[(125+2000)×1876]:2=1993250

(1+5+5+.....+65)