1—2/1X(1+2)—3/(1+2)X(1+2+3)—4/(1+2+3)X(1+2+3+4)—…—10/(1+2+3+…+9）X（1+2+3+…+10）

1+2+3+……+(n-1)=n(n-1)/2
1+2+3+……+n=n(n+1)/2
n/{[1+2+3+……+(n-1)][1+2+3+……+n]}
=n/[n(n-1)/2*n(n+1)/2]
=4/[(n-1)n(n+1)]
=2[1/(n-1)-2/n+1/(n+1)]
1—2/1X(1+2)—3/(1+2)X(1+2+3)—4/(1+2+3)X(1+2+3+4)—…—10/(1+2+3+…+9）X（1+2+3+…+10）
=1-2(1/1-2/2+1/3)-2(1/2-2/3+1/4)-2(1/3-2/4+1/5)-……-2(1/9-2/10+1/11)
=1-2[1/1-2/2+1/2+1/11-2/11+1/12]
=1-2[1/2-1/11+1/12]
=1-2[65/132]
=1-65/66
=1/66

像（1+2+3+…+10）这个是在分子还是分母上呀？

原式=1-2/1+2-1/1+2 + 1/1+2+3-1/1+2+3…-1/1+2+3+4+…10 +1/1+2+3+…10
=0

原式=1-(3-1)/1*3-(6-3)/3*6-(10-6)/6*10-...-(55-45)/45*55
=1-(1-1/3)-(1/3-1/6)-(1/6-1/10)-...-(1/45-1/55)
=1-1+1/3-1/3+1/6-1/6+1/10-...-1/45+1/55
=1/55

(n+1)/(1+2+....+(n+1))*(1+2+....+n)=(n+1)/{1/2*(n+1)(n+2)*[1/2*n*(n+1)]}=4*1/[n*(n+1)*(n+2)]=2*{1/[n*(n+1)]-1/[(n+2)*(n+1)]}

所以最后答案为：1/55