直线ln:y=-n/(n+1)*x+√2/(n+1)令y=0,得x=√2/n,ln与x轴交于An(√2/n,0)令x=0,得y=√2/(n+1),ln与y轴交于Bn(0,√2/(n+1))∴ln与两坐标围成的三角形面积Sn=1/2*|OAn|*|Bn}=1/[n(n+1)]∴S1+S2+S3+...+S2012=1/(1*2)+1/...