裂项相消：
1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+2011)(x+2012)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2011)-1/(x+2012) 中间的都消掉了
=1/x-1/(x+2012)
=(x+2012-x)/[x(x+2012)]
=2012/[x(x+2012)]