各项均为正数的数列{an},其前n项的和为Sn,且Sn=(根号S(n-1 )+根号a1)^2(n≥2),若bn=a(n+
各项均为正数的数列{an},其前n项的和为Sn,且Sn=(根号S(n-1 )+根号a1)^2(n≥2),若bn=a(n+1)/an+an/a(n+1),
且数列bn前n项的和为Tn,则Tn为
且数列bn前n项的和为Tn,则Tn为
赞 pqh163 幼苗
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根据条件,√S(n)-√S(n-1)=√a1,由此√S(n)-√S(1)=(n-1)√a1,S(n)=n^2*a1
bn=(S(n+1)-S(n))/(S(n)-S(n-1))+(S(n)-S(n-1))/(S(n+1)-S(n)),a1≠0
=((n+1)^2-n^2))/(n^2-(n-1)^2)+(n^2-(n-1)^2)/((n+1)^2-n^2)=(2n+1)/(2n-1)+(2n-1)/(2n+1)
=4+2/(2n-1)-2/(2n+1)
Σbk=4n+2(Σ1/(2k-1)-Σ1/(2k+1))=4n+2-2/(2n+1)
bn=(S(n+1)-S(n))/(S(n)-S(n-1))+(S(n)-S(n-1))/(S(n+1)-S(n)),a1≠0
=((n+1)^2-n^2))/(n^2-(n-1)^2)+(n^2-(n-1)^2)/((n+1)^2-n^2)=(2n+1)/(2n-1)+(2n-1)/(2n+1)
=4+2/(2n-1)-2/(2n+1)
Σbk=4n+2(Σ1/(2k-1)-Σ1/(2k+1))=4n+2-2/(2n+1)
1年前
9
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