问题 解答题
数列{an}的首项a1=1,前n项和为Sn,满足关系3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4…)
(1)求证:数列{an}为等比数列;
(2)设数列{an}的公比为f(t),作数列{bn},使b1=1,bn=f(
1
bn-1
),(n=2,3,4…),求bn
(3)求Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)的值.
答案

(1)证:∵3tSn-(2t+3)Sn-1=3t,3tSn+1-(2t+3)Sn=3t(n≥2),两式相减得3tan+1-(2t+3)an=0

又t>0

an+1
an
=
2t+3
3t
(n≥2),

又当n=2时,3tS2-(2t+3)S1=3t,

即3t(a1+a2)-(2t+3)a1=3t,得a2=

2t+3
3t

a2
a1
=
2t+3
3t

an+1
an
=
2t+3
3t
(n≥1),

∴{an}为等比数列

(2)由已知得,f(t)=

2t+3
3t

∴bn=f(

1
bn-1
)=
3+
2
bn-1
3
bn-1
=
2
3
+bn-1(n≥2,n∈N*).

∴{bn}是一个首项为1,公差为

2
3
的等差数列.

于是bn=

2
3
n+
1
3

(3)Tn=b1b2-b2b3+b3b4-…+b2n-1b2n-b2nb2n+1

=b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1-b2n+1)=-2(b2+b4+…+b2n

=-2d(b2+b4+…+b2n

=-2×

2
3
[
5n
3
+
n(n-1)
2
×
4
3
]

=-

8n2
9
-
4n
3

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