已知数列{an}的首项a1=
(Ⅰ)求证:数列{
(Ⅱ)记Sn=
|
(Ⅰ)证明:∵
=1 an+1
+2 3
,∴1 3an
-1=1 an+1
-1 3an
,1 3
∵
-≠0,∴1 a1
-1≠0(n∈N+),1 an
∴数列{
-1}为等比数列.1 an
(Ⅱ)由(Ⅰ)可求得
-1=1 an
×(2 3
)n-1,∴1 3
=2×(1 an
)n+11 3
Sn=
+1 a1
+…+1 a2
=n+2×(1 an
+1 3
+…+1 32
)=n+2•1 3n
=n+1-
-1 3 1 3n+1 1- 1 3
,1 3n
若Sn<100,则n+1-
<100,1 3n
∴nmax=99.