A. 1·x + 2·x + 3·x + ... + 2010·x = 1005 · 2011
x(1+2+3+...+2010)=1005*2011
1+2+3+...+2010=
2010*(2010+1)/2=
2010*2011/2=
1005*2011
x(1005*2011)=1005*2011
=> x=1
1·x - 2·x + 3·x - ... - 2010x + 2011x = 2012
x(1-2+3-4+........-2010+2011)=2012
1-2+3-4+........-2010+2011=
COnform Euler=1/4 - FOrmula asta se fol atunci cand tinde spre infinit.
TU VEI AVEA
-1+-1+-1+...+-1+2011=
-1*1005+2011
-1005+2011=1006
x*1006=2012
x=2012:1006
x=2
