分析:設(shè)數(shù)列的公比為q,當(dāng)q=1時(shí)則S
n=na
1,代入S
n,S
n+2,S
n+1,再根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性得證,當(dāng)≠1時(shí)把等比數(shù)列的求和公式
Sn=代入S
n,S
n+2,S
n+1,再根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性得證.
解答:證明:設(shè){a
n}的公比為q,由題設(shè)知a
1>0,q>0,
(1)當(dāng)q=1時(shí),S
n=na
1,從而
S
n•S
n+2-S
n+12=na
1(n+2)a
1-(n+1)
2a
12=-a
12<0.
(2)當(dāng)q≠1時(shí),
Sn=,從而
S
n•S
n+2-S
n+12=
-=-a
12q
n<0.
由(1)和(2)得S
n•S
n+2<S
n+12.
根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性,得log
0.5(S
n•S
n+2)>log
0.5S
n+12,
即
>log0. 5Sn+1.
點(diǎn)評(píng):本小題主要考查等比數(shù)列、對(duì)數(shù)、不等式等基礎(chǔ)知識(shí)以及邏輯推理能力,