已知数列{an}的通项公式an=n²,数列{bn}的首项b1=3,其前n项和为Sn,且满足关系式[a（n+1）+a（n+2）+……+a（2n）]/Sn=（7×n+1）/6①求{bn}的通项公式②求证：数列{2^（-bn）}是一个等比数列,若他

Question

已知数列{an}的通项公式an=n²,数列{bn}的首项b1=3,其前n项和为Sn,且满足关系式[a（n+1）+a（n+2）+……+a（2n）]/Sn = （7×n+1）/6
①求{bn}的通项公式
②求证：数列{2^ （-bn）}是一个等比数列,若他的前n项和Tn＞31/240,求n的范围.

Answer 1

1.
设an的前n项和为An
an=n^2
An=n(n+1)(2n+1)/6
a(n+1)+a(n+2)+……+a(2n)
=A(2n)-An
=2n(2n+1)(4n+1)/6-n(n+1)(2n+1)/6
=n(2n+1)(7n+1)/6
Sn=[a(n+1)+a(n+2)+……+a(2n)]/[(7n+1)/6]
=[n(2n+1)(7n+1)/6]/[(7n+1)/6]
=n(2n+1)
Bn=Sn-S(n-1)
=n(2n+1)-(n-1)(2n-1)
=4n-1
经验算B1=3也满足上式
2.
2^(-Bn)/2^(-B(n-1))
=2^[(-Bn)-(-B(n-1)]
=2^(-4)
{2^(-Bn)}是等比数列
2^(-Bn)=2^(-4n+1)
B1=2^(-3)
Tn=2^(-3)×(1-(2^(-4))^n)/(1-2^(-4))>31/240
(1-(2^(-4)^n)/(1-2^(-4))>31/30
2^4-2^(-4n+4)/(2^4-1)>31/30
16-2^(-4n+4)>31/2
2^(-4n+4)=2