已知数列{an}的首项a1=23,an+1=2anan+1,n=1,2,3,….令bn=1an-1.(Ⅰ)证明:数列{bn}是等比数列,

2025-05-13 13:18:58
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解答:(Ⅰ)证明:∵an+1=

2an
an+1

an+1an +an+1=2an,∴1+
1
an
=
2
an+1

1
2
+
1
2an
1
an+1
,∴
1
2
(
1
an
?1)=
1
an+1
?1

a1
2
3
,∴
1
a1
?1=
3
2
?1=
1
2

∵bn=
1
an
-1,∴{bn}是首项为
1
2
,公比为
1
2
的等比数列,
∴bn=
1
an
-1=(
1
2
)
n

(Ⅱ)解:∵cn=2n?bn=2n?(
1
2
n
∴Tn=2?
1
2
+4?
1
22
+6?
1
23
+…+2n?
1
2n
,①
1
2
Tn
=2?
1
22
+4?
1
23
+6?
1
24
+…+2n?
1
2n+1 
,②
∴①-②,得
1
2
Tn
=1+
1
2
+
1
22 
+…+
1
2n?1
-
n
2n 

=
1?
1
2n
1?
1
2
-
n
2n

=2-
n+2
2n

Tn =4-
n+2
2n?1