Question

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

Answer 1

CAND SE SCD CELE DOUA PARANTEZE SE OBTINE O SUMA DE 1 ( PT CA 2-1=1; 4-3=1....2012-2011=1)=> S=1+1+1+1+...1 AVEM 1006 TERMENI => S=1006

Answer 2

$\displaystyle (2+4+...+2012)-(1+3+...+2011) \\ \\ 2+4+...+2012=2(1+2+...+1006)=2 \times \frac{1006(1006+1)}{2} = \\ \\ =2 \times \frac{1006 \times 1007}{2} =\not2 \times \frac{1013042}{\not2} =1013042$

$\displaystyle 1+3+...+2011=1+2+3+...+2011-(2+4+...+2010)= \\ \\ = \frac{2011(2011+1)}{2} -2(1+2+...+1005)= \\ \\ = \frac{2011 \times 2012}{2} -2 \times \frac{1005(1005+1)}{2} = \frac{4046132}{2} -2 \times \frac{1005 \times 1006}{2} = \\ \\ =2023066-\not2 \times \frac{1011030}{\not2} =2023066-1011030=1012036$

$(2+4+...+2012)-(1+3+...+2011)= \\ \\ =1013042-1012036=1006$