解：当n≥2时：Sn=1/3n(n²-1)Sn-1=1/3(n-1)[(n-1)²-1]相减得an=1/3[n³-(n-1)³-1]=1/3[n²+(n-1)²+n(n-1)]=n²-n= n(n-1)∴1/an=1/n(n-1)=1/(n-1)-1/n1/a2+1/a3+,.......+1/an+1=1-1/2+1/2-1/3+...+1/(n-1)-1/n=1-1/n
