a1+a2+a3...an=n*n*na1+a2+a3...a(n-1)=(n-1)*(n-1)*(n-1)两式相减得an=3n^2-3n+1于是1/(an-1)=1/3*n*(n-1)=1/3[1/(n-1)-1/n)]所以1/a2-1+(1/a3-1)+.1/a100-1=1/3（1-1/2+1/2-1/3+1/3-1/4+...-1/100）=33/100...