【答案】
分析:(Ⅰ)由點(diǎn)(1,0)在函數(shù)f(x)上,可以得到關(guān)系式a
n+1=
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a
n,且a
1=
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,在利用求解等比數(shù)列通項(xiàng)公式的方法求解即可.
(Ⅱ)由(Ⅰ)得到數(shù)列a
n為等比數(shù)列,將a
n的通項(xiàng)公式代入b
n=log
2a
2n-1中即可求得b
n的通項(xiàng)公式,進(jìn)而求出數(shù)列{b
n}的前n項(xiàng)和T
n.
解答:解:(Ⅰ)由已知得f(1)=
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a
n-a
n+1=0,解得a
n+1=
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a
n,
∵
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所以數(shù)列{a
n}是首項(xiàng)為
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、公比為
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的等比數(shù)列.
所以通項(xiàng)公式
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.
(Ⅱ)由b
n=log
2a
2n-1=log
2a
2n-1=1-2n
所以數(shù)列{b
n}的前n項(xiàng)和T
n=(-1)+(-3)+(-5)+…+(1-2n)=-n
2.
點(diǎn)評:本題主要考查了數(shù)列通項(xiàng)公式和前n項(xiàng)和的求解問題,解題時注意整體思想和轉(zhuǎn)化思想的運(yùn)用,平時多練習(xí),注意解題步驟,才能夠做到舉一反三,屬于中檔題.