math problem please1+2分之一加上1+2+3分之一加上1+2+3+4分之一一直加到1+2+3+．．．＋100等于几?

1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
=2/6+2/12+2/20+...+2/(101*100)
=2(1/(2*3)+1/(3*4)+1/(4*5)+...+1/(100*101))
=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/100-1/101_
=2(1/2-1/101)
=1-2/101
=99/101

1+2+3+...+n=(1+n)*n/2,
1/(1+2+3+...+n)=1/[n(n+1)/2]=2/[n(n+1)],
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+99+100)
=2/(2*3)+2/(3*4)+1/(4*5)+...+1/(100*101)
=2*[1/(2*3)+1/(3*4)+1/(4*5)+...+1/(100*101)]
=2[(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+...+(1/100)-(1/101)]
=2[(1/2)-(1/101)]
=2[(101-2)/202]
=99/101.

原式=99/101
解题步骤：
∵1／（n+1）n=1/n-1/(n+1) 这是解题关键
1/(1+2)=1/(1+2)×2÷2=2/(1+2)×2
1/(1+2+3)=1/(1+3)×3÷2=2/(1+3)×3
1/(1+2+3+......+n)=1/(1+n)n÷2=2/(1+n)n=2×[1/(n+1)n]=2[1/n-1/(n+1)]
∴原式=2/(1+2)×2+2/(1+3)×3......+2/(1+100)×100
=2{1/(1+2)×2+1/(1+3)×3......+1/(1+100)×100}
=2{1/2-1/3+1/3-1/4+.-......+1/100-1/101}
=2{1/2-1/101}
=1-2/101
=99/101

首先求出通项公式:An=[1/(1+2+3+...+n)]-1=[2/n(n+1)]-1
所以Sn=中间自己算=[2n/(n+1)]-1
当n=100时,Sn=(200/101)-1=99/101
这类问题主要的思想都是先求出通项.

1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+100)=1/3+1/6+1/10+...+1/5050=2/6+2/12+2/20+...+2/10100=2/(2*3)+2/(3*4)+2/(4*5)+...+2/(100*101)=1-2/3+2/3-2/4+2/4-2/5+...+2/100-2/101=1-2/101=99/101