Question

在数列｛an｝中，an=1/(n+1)+2/(n+1)+···+n/(n+1),bn=2/[an·a(n+1)]，求数列｛bn｝的前n项和。

如题，要过程！谢谢！

Answer 1

an=1/(n+1)+2/(n+1)+...+n/(n+1)
=(1+n)*n/(n+1)2=n/2
an+1=(n+1)/2
bn=2/[n/2*(n+1)/2]=8/[n(n+1)]
=8[1/n-1/(n+1)]
b1+b2+b3+...bn=8[1-1/2+1/2-1/3+1/3-1/4+
...1/n-1/(n+1)
=8[n/(n+1)]
=8n/(n+1)

祝您学习愉快

Answer 2

b1+b2+b3+...bn=8[1-1/2+1/2-1/3+1/3-1/4+
...1/n-1/(n+1)
=8[n/(n+1)]
=8n/(n+1)