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)
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
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,
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,
数学人气:376 ℃时间:2020-02-02 18:47:16
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