【答案】
分析:(1)利用式子(3-m)S
n+2ma
n=m+3求出(3-m)S
n+1+2ma
n+1=m+3,相減得到
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為常蘇,即可得證.
(2)先求出b
1=1,再根據(jù)題意得到數(shù)列{b
n}的表達(dá)式,構(gòu)造新的數(shù)列,求出新蘇烈的表達(dá)式,進(jìn)而求出數(shù)列{b
n}的表達(dá)式.
解答:解:(1)證明:有(3-m)S
n+2ma
n=m+3,得(3-m)S
n+1+2ma
n+1=m+3,(2分)
兩式相減,得(3+m)a
n+1=2ma
n(m≠3)(4分)
∴
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為常數(shù),∴{a
n}是等比數(shù)列(5分)
(2)由(3-m)a
1+2ma
1=m+3,得(m+3)a
1=m+3,∵m≠-3∴a
1=1,b
1=1,(6分)
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,且n≥2時,
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,(7分)
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,(9分)
∴
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是1為首項(xiàng)
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為公差的等差數(shù)列,(10分)
∴
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(11分)
∴
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(12分)
點(diǎn)評:此題主要考查等比數(shù)列的這鞥名以及構(gòu)造新數(shù)列求解數(shù)列表達(dá)式的方法.