已知直线y=−(n+1)n+2x+1n+2（n为正整数）与坐标轴围成的三角形的面积为Sn，则S1+S2+S3+…+S2012=．

Question

已知直线y=(−(n+1))/(n+2)x+(1)/(n+2)（n为正整数）与坐标轴围成的三角形的面积为S_n，则S₁+S₂+S₃+…+S₂₀₁₂=______．

Answer 1

令x=0，则y=(1)/(n+2)，
令y=0，则-(n+1)/(n+2)x+(1)/(n+2)=0，
解得x=(1)/(n+1)，
所以，S_n=(1)/(2)•(1)/(n+1)•(1)/(n+2)=(1)/(2)（(1)/(n+1)-(1)/(n+2)），
所以，S₁+S₂+S₃+…+S₂₀₁₂=(1)/(2)（(1)/(2)-(1)/(3)+(1)/(3)-(1)/(4)+(1)/(4)-(1)/(5)+…+(1)/(2013)-(1)/(2014)）=(1)/(2)（(1)/(2)-(1)/(2014)）=(503)/(2014)．
故答案为：(503)/(2014)．