已知直线y=[-(n+1)/(n+2)]x+[1/(n+2)](n为正整数)与两坐标轴围成的三角形面积为Sn,则S1+S2+S3.S2012=

已知直线y=[-(n+1)/(n+2)]x+[1/(n+2)](n为正整数)与两坐标轴围成的三角形面积为Sn,则S1+S2+S3.S2012=
答案是这样的令x=0
y=1/(n+2)
令y=0
x=1/(n+1)
∴Sn=1/2*1/(n+2)*1/(n+1)
=1/2*1/[(n+1)*(n+2)]
=1/2*[1/(n+1)-1/(n+2)]
裂项求和
S1+S2+S3.S2012
=1/2*(1/2-1/3+1/3-1/4+1/4-1/5+.+1/2012-1/2013+1/2013-1/2014)
=1/2*(1/2-1/2014)
=1/2*1006/2014
=503/2014
请问
1/2*1/[(n+1)*(n+2)]
=1/2*[1/(n+1)-1/(n+2)]
是怎么变的,为什么后面变成了减号