Sn=n^2,令bn=1/anan+1,Tn是数列bn的前n项和,试证明Tn
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Sn=n^2,令bn=1/anan+1,Tn是数列bn的前n项和,试证明Tn
Sn=n^2
n>=2时,
An=Sn-S(n-1)=n^2-(n-1)^2=2n-1
n=1时,A1=S1=1也满足上式
Bn=1/AnA(n+1)
=1/[(2n-1)(2n+1)]
=(1/2)[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]
=(1/2)[1/(2n-1)-1/(2n+1)]
Tn=B1+B2+……+Bn
=(1/2)[(1/1-1/3)+(1/3-1/5)+……+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
n>=2时,
An=Sn-S(n-1)=n^2-(n-1)^2=2n-1
n=1时,A1=S1=1也满足上式
Bn=1/AnA(n+1)
=1/[(2n-1)(2n+1)]
=(1/2)[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]
=(1/2)[1/(2n-1)-1/(2n+1)]
Tn=B1+B2+……+Bn
=(1/2)[(1/1-1/3)+(1/3-1/5)+……+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
Sn=n^2,令bn=1/anan+1,Tn是数列bn的前n项和,试证明Tn
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