Sn=n(n+1)(n+2)/3 Sn-1=(n-1)n(n+1)/3
an=Sn-Sn-1=n(n+1)(n+2-n+1)/3=n(n+1)
1/an=1/[n(n+1)]=[(n+1)-n]/[n(n+1)]=1/n - 1/(n+1)
数列an分之一的前n项和=1 - 1/2 + 1/2 -1/3+……+1/n - 1/(n+1)=1-1/(n+1)=n/(n+1)
Sn=n(n+1)(n+2)/3 Sn-1=(n-1)n(n+1)/3
an=Sn-Sn-1=n(n+1)(n+2-n+1)/3=n(n+1)
1/an=1/[n(n+1)]=[(n+1)-n]/[n(n+1)]=1/n - 1/(n+1)
数列an分之一的前n项和=1 - 1/2 + 1/2 -1/3+……+1/n - 1/(n+1)=1-1/(n+1)=n/(n+1)