裂项相消求和:数列{an}中,an=1/(n+1)+2/(n+1)+3/(n+1)+……+n/(n+1),bn=2/(a
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裂项相消求和:数列{an}中,an=1/(n+1)+2/(n+1)+3/(n+1)+……+n/(n+1),bn=2/(an*a(n+1))
,求数列{bn}的前n项
,求数列{bn}的前n项
![裂项相消求和:数列{an}中,an=1/(n+1)+2/(n+1)+3/(n+1)+……+n/(n+1),bn=2/(a](/uploads/image/z/17472137-41-7.jpg?t=%E8%A3%82%E9%A1%B9%E7%9B%B8%E6%B6%88%E6%B1%82%E5%92%8C%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Can%3D1%2F%28n%2B1%29%2B2%2F%28n%2B1%29%2B3%2F%28n%2B1%29%2B%E2%80%A6%E2%80%A6%2Bn%2F%28n%2B1%29%2Cbn%3D2%2F%28a)
an=1/(n+1)+2/(n+1)+3/(n+1)+……+n/(n+1)=(1+2+3+.+(n+1))/(n+1)
=((n+2)(n+1)/2)/(n+1)=(n+2)/2
所以a(n+1)=(n+3)/2
所以bn=2/(an*a(n+1))=2/((n+2)(n+3)/4)=8/(n+2)(n+3)=8(1/(n+2)-1/(n+3))
所以Sn=8(1/3-1/4+1/4-1/5+.+1/(n+2)-1/(n+3))=8(1/3-1/(n+3))=8n/3(n+3)
=((n+2)(n+1)/2)/(n+1)=(n+2)/2
所以a(n+1)=(n+3)/2
所以bn=2/(an*a(n+1))=2/((n+2)(n+3)/4)=8/(n+2)(n+3)=8(1/(n+2)-1/(n+3))
所以Sn=8(1/3-1/4+1/4-1/5+.+1/(n+2)-1/(n+3))=8(1/3-1/(n+3))=8n/3(n+3)
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