Question

1+2+3+........+44=
1+2+3+........+167=
2+4+6+...…...+244=
3+6+9+.........+2001=
5+10+15+.........+200=
7+14+21+.........+3157=

Answer 1

       Folosim formula lui Gauss...
       1+2+3+...+n = n*(n+1) / 2 
1+2+3+...+44 = 44*45 / 2 =990 ;
1+2+3+...+167=167*168 / 2 =14028 ;
2+4+6+...+244=2*(1+2+3+...+122)=2* 122*123 / 2 =15006 ;
=3*(1+2+3+...+667)=3*667*668 / 2=668334 ;
=5*(1+2+3+...+40)=5*40*41 /2 =4100 ;
=7*(1+2+3+...+451)=7*451*452 / 2=713482 ;

Answer 2

sume Gauss.

Suma Gauss (numerele sunt consecutive si incep de la 1)  care are formula de rezolvare:[n(n+1)]/2  unde n este ultimul numar al sirului:

a) 1+2+3+........+44=[n(n+1)]/2 =[44(44+1)]:2=(44×45):2=990

b. 1+2+3+........+167=[n(n+1)]/2 =[167(167+1)]:2=14028

c) 2+4+6+...…...+244=2(1+2+3+.....+122)=2×{[122(122+1)]:2}=2×7503=15006

 d) 3+6+9+.........+2001=3(1+2+3+....+667)=3×{[667(667+1)]:2}=
=3×222778=668334

e) 5+10+15+.........+200=5(1+2+3+....40)=5×{[40(40+1)]:2} =5×820=4100

f) 7+14+21+.........+3157= 7(1+2+3+....+451)=7×{[451(451+1)]:2}
=7×101926=713482