题目内容

在数列{an}中,a1=1,an+1
an
=8

(Ⅰ)求a2,a3
(Ⅱ)设bn=log2an,求证:{bn-2}为等比数列;
(Ⅲ)求{an}的前n项积Tn
(Ⅰ)∵a2
a1
=8,a1=1

∴a2=8.
a3
a2
=8,a1=8

a3=2
2

(Ⅱ)证明:∵
bn+1-2
bn-2
=
log2an+1-2
log2an-2

=
log2
8
an
-2
log2an-2
=
3-
1
2
log2an-2
log2an-2

1
2
×
2-log2an
log2an-2
=-
1
2

∴{bn-2}为等比数列,首项为b1-2,即为-2,其公比为-
1
2

(Ⅲ)设数列{bn-2}的前n项和为Sn
Sn=
-2(1-(-
1
2
)
n
)
1+
1
2
=b1+b2+b3+…+bn-2n=log2a1+log2a2+…log2an-2n
=log2Tn-2n

log2Tn=
4
3
[(-
1
2
)n-1]+2n

Tn=2
4
3
[(-
1
2
)
n
-1]+2n
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