问题 解答题
已知数列{an}的首项a1=
3
5
an+1=
3an
2an+1
,其中n∈N+
(Ⅰ)求证:数列{
1
an
-1
}为等比数列;
(Ⅱ)记Sn=
1
a1
+
1
a2
+…+
1
an
,若Sn<100,求最大的正整数n.
答案

(Ⅰ)证明:∵

1
an+1
=
2
3
+
1
3an
,∴
1
an+1
-1=
1
3an
-
1
3

1
a1
-≠0,∴
1
an
-1≠0(n
∈N+),

∴数列{

1
an
-1}为等比数列.

(Ⅱ)由(Ⅰ)可求得

1
an
-1=
2
3
×(
1
3
)
n-1
,∴
1
an
=2×(
1
3
)n+1

Sn=

1
a1
+
1
a2
+…+
1
an
=n+2×(
1
3
+
1
32
+…+
1
3n
)
=n+2•
1
3
-
1
3n+1
1-
1
3
=n+1-
1
3n

若Sn<100,则n+1-

1
3n
<100,

∴nmax=99.

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