【答案】
分析:(1)依題意可求得a
2的值,進(jìn)而求得
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的值,進(jìn)而看當(dāng)n≥2時(shí),根據(jù)a
n=S
n-S
n-1求得
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判斷出數(shù)列為等比數(shù)列,進(jìn)而根據(jù)等比數(shù)列的性質(zhì)求得a
n,進(jìn)而分別表示出lga
n和lga
n+1,根據(jù)lga
n+1-lga
n=1,判斷出lga
n}n∈N
*是等差數(shù)列.
(2)根據(jù)(1)中求得a
n利用裂項(xiàng)法求得T
n,進(jìn)而根據(jù)3-
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≥
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,進(jìn)而根據(jù)
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求得m的范圍.判斷出m的最大正整數(shù).
解答:解:(1)依題意,
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,
當(dāng)n≥2時(shí),a
n=9S
n-1+10①又a
n+1=9S
n+10②
②-①整理得:
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為等比數(shù)列,
且a
n=a
1q
n-1=10
n,∴l(xiāng)ga
n=n∴l(xiāng)ga
n+1-lga
n=(n+1)-n=1,
即{lga
n}n∈N
*是等差數(shù)列.
(2)由(1)知,
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=
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∴
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,
依題意有
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,
故所求最大正整數(shù)m的值為5.
點(diǎn)評(píng):本題主要考查了等差數(shù)列和等比數(shù)列的性質(zhì).考查了學(xué)生對(duì)數(shù)列基礎(chǔ)知識(shí)的綜合把握.