1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3...+n)
=1+1*2/(2*3)+1*2/(3*4)+...+1*2/[n*(1+n)]
=2[1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/(n+1)]
=2[1/2+1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)
我给你求出了通项公式 接下去你应该会了把