数列{an}的前N项和为Sn,数列{bn}的前n项的和Tn,{bn}为等差数列且各项均为正数,a1=1,an+1=2Sn+1(n属于正整数),T3=15.(1)求证:数列{an}是等比数列(2)若a1+b1,a2+b2,a3+b3成等比数列,求Tn
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![数列{an}的前N项和为Sn,数列{bn}的前n项的和Tn,{bn}为等差数列且各项均为正数,a1=1,an+1=2Sn+1(n属于正整数),T3=15.(1)求证:数列{an}是等比数列(2)若a1+b1,a2+b2,a3+b3成等比数列,求Tn](/uploads/image/z/3022197-69-7.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8DN%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8CTn%2C%7Bbn%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E4%B8%94%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2Ca1%3D1%2Can%2B1%3D2Sn%2B1%EF%BC%88n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%2CT3%3D15.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E8%8B%A5a1%2Bb1%2Ca2%2Bb2%2Ca3%2Bb3%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E6%B1%82Tn)
数列{an}的前N项和为Sn,数列{bn}的前n项的和Tn,{bn}为等差数列且各项均为正数,a1=1,an+1=2Sn+1(n属于正整数),T3=15.(1)求证:数列{an}是等比数列(2)若a1+b1,a2+b2,a3+b3成等比数列,求Tn
数列{an}的前N项和为Sn,数列{bn}的前n项的和Tn,{bn}为等差数列且各项均为正数,a1=1,an+1=2Sn+1(n属于正整数),T3=15.
(1)求证:数列{an}是等比数列
(2)若a1+b1,a2+b2,a3+b3成等比数列,求Tn
数列{an}的前N项和为Sn,数列{bn}的前n项的和Tn,{bn}为等差数列且各项均为正数,a1=1,an+1=2Sn+1(n属于正整数),T3=15.(1)求证:数列{an}是等比数列(2)若a1+b1,a2+b2,a3+b3成等比数列,求Tn
(2)
{bn}为等差数列,公差为d
则b1+b3=2b2
Tn=b1+b3+b2=3b2=15,则b2=5
b1=5-d,b2=5+d
a1+b1,a2+b2,a3+b3成等比数列
则(a2+b2)^2=(a1+b1)(a3+b3)
(5+3)^2=[1+(5-d)][9+(5+d)]
解得,d=2或-10({bn}的各项均为正,故舍去)
bn=2n+1
Tn=n[3+(2n+1)]/2=n(n+2) =n^2+2n
a(n+1)=2S(n+1)
下标吗?
1)
a(n+1)=2Sn+1--(1)
an=2S(n-1)+1--(2)
(1)-(2),得
a(n+1)-an=2Sn-2S(n-1)=2an
得a(n+1)=3an
所以{an}为等比数列,公比为3
an=3^(n-1)
(2)
{bn}为等差数列,公差为d
则b1+b3=2b2
Tn=...
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1)
a(n+1)=2Sn+1--(1)
an=2S(n-1)+1--(2)
(1)-(2),得
a(n+1)-an=2Sn-2S(n-1)=2an
得a(n+1)=3an
所以{an}为等比数列,公比为3
an=3^(n-1)
(2)
{bn}为等差数列,公差为d
则b1+b3=2b2
Tn=b1+b3+b2=3b2=15,则b2=5
b1=5-d,b2=5+d
a1+b1,a2+b2,a3+b3成等比数列
则(a2+b2)^2=(a1+b1)(a3+b3)
(5+3)^2=[1+(5-d)][9+(5+d)]
解得,d=2或-10({bn}的各项均为正,故舍去)
bn=2n+1
Tn=n[3+(2n+1)]/2=n(n+2)
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