Question

Sa se calculeze (2+4+...+2012)-(1+3+...+2011)

Answer 1

(2+4+...+2012)=2(1+2+3+...+1006)= 2* (1+1006)*1006/2=  1007*1006

(1+3+5+....+2011)=(1+2011)*1005/2= 2012*1005/2= 1006*1005

1007*1006-1006*1005= 1006(1007-1005)= 1006*2=2012

Answer 2

(2 + 4 + .... + 2012) - (1 + 3 +.....+ 2011) =

= -1 + 2 - 3 + 4 - 5 + 6 -........- 2011 + 2012 =
Avem 2012 termeni

Grupam cate 2 si obtinem 2012 / 2 = 1006 grupe

= (-1 + 2) + (-3 + 4) + (-5 + 6) + .......+ (-2011 + 2012) = 

= ( 1 ) + ( 1 ) + ( 1 ) +.............+( 1 ) =   ( Este o adunare repetata de 1006 termeni)

1006 * 1 = 1006