在数列{an}中,an=1/(n+1)+2/(n+1)+...+n/(n+1),bn=2/[an*(an+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)

解： 因为1+2+3+……+（n-1）+n=(n+1)n/2

an=1/(n+1)+2/(n+1)+...+n/(n+1)=[1+2+3+……+（n-1）+n]/(n+1)=[(n+1)n/2]/(n+1)=n/2

bn=2/[an*(an+1)]=4/[n(n+1)]

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

如果要求{bn}的和的话要讨论的,记bn的和为Bn
如果n为偶数的时候，Bn=(-1pf+2pf)+(-3pf+4pf)+...+(-(n-1)pf+npf)=1+2+3+4+...+n-1+n=n*(n+1)/2
如果n为奇数的时候，Bn=B（n-1）+bn=(n-1)*n/2-npf=-n*(n+1)/2,懂了吗？