已知曲線C:y=x2(x>0),過C上的點A1(1,1)作曲線C的切線l1交x軸于點B1,再過B1作y軸的平行線交曲線C于點A2,再過A2作曲線C的切線l2交x軸于點B2,再過B2作y軸的平行線交曲線C于點A&3,…,依次作下去,記點An的橫坐標為an(n∈N*).
(Ⅰ)求數(shù)列{an}的通項公式;
(Ⅱ)記bn=(8-2n)an,設(shè)數(shù)列{bn}的前n項和為Tn,求證:0<Tn≤4.
【答案】
分析:(I)由y'=2x(x>0).知切線l
n的方程為y-a
n2=2a
n(x-a
n).所以
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.依題意點A
n+1在直線
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上,所以數(shù)列{a
n}是1為首項,
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為公比的等比數(shù)列.由此能求出數(shù)列{a
n}的通項公式.
(Ⅱ)由
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,知
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.由錯位相減法能導(dǎo)出
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,n≥2時,
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.由n≥2時,T
n≤T
n-1,知T
n≤T
n-1≤…≤T
2,由此能夠證明0<T
n≤4.
解答:解(I)∵y'=2x(x>0).∴曲線C在點A
n(a
n,a
n2)處的切線l
n的斜率為k
n=2a
n.
∴切線l
n的方程為y-a
n2=2a
n(x-a
n).(2分)
令y
=0得
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,
∴
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.
依題意點A
n+1在直線
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上,
∴
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又a
1=1.(4分)
∴數(shù)列{a
n}是1為首項,
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為公比的等比數(shù)列.
∴
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.(5分)
(Ⅱ)由已知
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.
∴
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.①
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.②
①-②得
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=
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=
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.(9分)
∴
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(10分)
又n≥2時,
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.
又當(dāng)n≥2時,T
n≤T
n-1.
∴T
n≤T
n-1≤…≤T
2.
∴當(dāng)n=2時,T
1=T
2=4.
∴(T
n)
max=T
2=4,∴T
n≤4.(13分)
綜上0<T
n≤4.(14分)
點評:本題考查通項公式的求法和求證:0<T
n≤4.解題時要認真審題,仔細解答,注意合理地進行等價轉(zhuǎn)化.