Question

在数列{An}中，An=1/(n+1)+2/(n+1)+…+n/(n+1),又bn=2/anan+1

在数列{An}中，An=1/(n+1)+2/(n+1)+…+n/(n+1),又bn=2/（anan+1），求bn的前n项和，急求！！！

Answer 1

an=1/(n+1)+2/(n+1)+...+n/(n+1)=(1+2+...+n)/(n+1)=[n(n+1)/2]/(n+1)=n/2
bn=2/[ana(n+1)]=2[(n/2)(n+1)/2]=8/[n(n+1)]=8[1/n-1/(n+1)]
Tn=b1+b2+...+bn
=8[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=8[1-1/(n+1)]
=8n/(n+1)

Answer 2

an=1/(n+1)+2/(n+1)+…+n/(n+1)=n(n+1)/2(n+1)=n/2, A(n+1)=(n+1)/2
则bn=8/n(n+1)=8(1/n-1/(n+1))
bn的前n项和为：8(1-1/2+1/2-1/3+1/3-1/4......-1/n+1/n-1/(n+1))=8(1-1/(n+1))