Question

regula lui gauss:
s=2+4+6+...+78+80
s=3+6+9+..+147+150
s=0+5+10+...+145+150
s=11+22+33+...242+253
s=33+34+35+...+79+80
s=7+8+9+...+45+46

Answer 1

Regula lui Gauss se aplica pentru suma numerelor consecutive S= [n*(n+1)]:2
Dam factor comun: 2(1+2+3+...+39+40)= 2[(40*41):2]= 2*(1640:2)=2*820=1640
3(1+2+3+....+49+50)= 3 [(50*51):2]= 3*(2550:2)= 3*1275=3825
5(1+2+........+29+30) = 5[(30*31):2]= 5*(930:2)=5*465=2325
11(1+2+3+....+22+23) = 11*[(23*24):2]= 11*(552:2)=11*276=3036
33+34+35+.......+79+80= [(80*81):2] - [(32*33):2]= 6480:2-1056:2=3240-528=2712
Observam ca la ex anterior lipsesc primi 32 de termeni din suma si am calculat suma pentru 80 de termeni scazand ce lipsea din ex adica suma primilor 32 de termeni, la fel procedam si pentru 7+8+9+.....+45+46 = [(46*47):2]- [(6*7):2]= 2162:2-42:2=1081-21=1060