题目内容

已知数列{an}为等比数列,a2=6,a5=162.
(1)求数列{an}的通项公式;
(2)设Sn是数列{an}的前n项和,证明
SnSn+2
S2n+1
≤1
(1)设等比数列{an}的公比为q,则a2=a1q,a5=a1q4
依题意,得方程组
a1q=6
a1q4=162

解此方程组,得a1=2,q=3.
故数列{an}的通项公式为an=2•3n-1
(2)Sn=
2(1-3n)
1-3
=3n-1

SnSn+2
S2n+1
=
32n+2-(3n+3n+2)+1
32n+2-2•3n+1+1
32n+2-2
3n3n+2
+1
32n+2-2•3n+1+1
=1

SnSn+2
S2n+1
≤1
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