【答案】
分析:(1)設(shè)等差數(shù)列{log
2(a
n-1)}的公差為d.根據(jù)a
1和a
3的值求得d,進(jìn)而根據(jù)等差數(shù)列的通項(xiàng)公式求得數(shù)列{log
2(a
n-1)}的通項(xiàng)公式,進(jìn)而求得a
n.
(2)把(1)中求得的a
n代入
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+
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+…+
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中,進(jìn)而根據(jù)等比數(shù)列的求和公式求得
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+
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+…+
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=1-
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原式得證.
解答:(I)解:設(shè)等差數(shù)列{log
2(a
n-1)}的公差為d.
由a
1=3,a
3=9得2(log
22+d)=log
22+log
28,即d=1.
所以log
2(a
n-1)=1+(n-1)×1=n,即a
n=2
n+1.
(II)證明:因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182220453877649/SYS201310241822204538776017_DA/7.png">=
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=
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,
所以
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+
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+…+
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=
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+
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+
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+…+
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=
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=1-
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<1,
即得證.
點(diǎn)評(píng):本題主要考查了等差數(shù)列的通項(xiàng)公式.屬基礎(chǔ)題.