解题思路:(1)由已知条件推导出
3
S
1
=
a
2
+(−2
)
3
−6
,S1=a1=2,由此能求出a2.
(2)由
3
S
n
=
a
n+1
+(−2
)
n+2
−6
,
3
S
n−1
=
a
n
+(−2
)
n+1
−6
两式相减得数列
{
a
n
(−2)
n
−1}
是首项为-2,公比为-2的等比数列,由此能求出数列{an}的通项公式.
(3)当m∈N*时,
1
a
2m
+
1
a
2m+1
=
1
4
2m
+
(−2)
2m
+
1
4
2m+1
+
(−2)
2m+1
<
5
4
2m+1
,由此进行分类讨论,能证明对一切正整数n,有[1