Question

(2+4+6+...+20)-(1+3+5+...19)

Answer 1

(2+4+6+...+20)-(1+3+5+...19) =

= 2×(1+2+3+..+10) - (2×0+1+2×1+1+2×2+1+...+2×9+1) =

= 2×10×(10+1):2 - 2×(0+1+2+...+9 + 1+1+1...+1 de 10 ori) =

= 2×5×11 - 2 × 9×10:2 + 1×10 =

= 110 - 2×9×5 + 10 =

= 110 - 90 + 10 =

= 110 - 100 =

= 10

Answer 2

S1 = 2 + 4 + 6 + ...... + 20
S1 = 20 + 18 + .............+ 2
___________________(+)
2S1 = 22+22+22+......+22
nr = (20-2):2 + 1 = 18:2+1 = 10
2S1= 22*10|:2
S1= 10*11 = 110

S2 = 1 + 3 + 5 + ...... + 19
S2 = 19 + 17 + ................+ 1
___________________(+)
2S2 = 20+20+20+......+20
nr = (19-1):2 + 1 = 18:2+1 = 10
2S2= 20*10|:2
S2= 10*10 = 100

S1 - S2 = 110 -100 = 10