Question 1050865
{{{1-2+3-4+5-6}}}+...+{{{2013}}}....take a look at signs:  positive, negative, positive,negative..
{{{odd}}} integers are positive and {{{even}}} integers are negative

there is {{{2013}}} integers in all
since first and last {{{odd}}} integers, means there is {{{one}}}{{{ even}}} integer {{{less}}} than odd

since {{{2013}}} is not divisible by {{{2}}} we can write it as {{{1007+1006 }}} where {{{1007}}} are odd integers and  {{{1006}}} even integer
since even integers are negative



SUM OF EVEN NUMBERS: {{{n[e](n[e]+1)}}}
SUM OF ODD NUMBERS: {{{n[o]^2}}}

since {{{even}}} integers are negative, we subtract the sum of even from the sum of odd integers 
 
{{{n[o]^2-n[e](n[e]+1)}}}...........{{{n[e]=1006}}} and {{{n[o]=1007}}} and we have
={{{ 1007^2-1006(1006+1)}}}
={{{1007^2-1006(1006+1)}}}
={{{1014049-1006^2-1006}}}
={{{1014049  -1012036-1006}}}
={{{1014049  -1013042}}}
={{{1007}}}