【答案】
分析:由題意再根據(jù)韋達(dá)定理列出a
n,a
n+1和b
n三者的關(guān)系式,再進(jìn)行變形求出數(shù)列{a
n}和{b
n}的特點,對數(shù)列分組求和,再由0<|c|<1求極限不等式,最后求出c的值.
解答:解:∵對任意n∈N*,a
n與a
n+1恰為方程x
2-b
nx+c
n=0的兩根,
∴a
n+a
n+1=b
n,a
n•a
n+1=c
n∴
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=
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=
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=c.
∵a
1=1,∴a
1•a
2=a
2=c.
∴a
1,a
3,a
5,…,a
2n-1,構(gòu)成首項為1,公比為c的等比數(shù)列,
a
2,a
4,a
6,…,a
2n,構(gòu)成首項為c,公比為c的等比數(shù)列.
又∵任意n∈N*,a
n+a
n+1=b
n恒成立.
∴
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=
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=c.又b
1=a
1+a
2=1+c,b
2=a
2+a
3=2c,
∴b
1,b
3,b
5,…,b
2n-1,構(gòu)成首項為1+c,公比為c的等比數(shù)列,
b
2,b
4,b
6,…,b
2n,構(gòu)成首項為2c,公比為c的等比數(shù)列,
∵0<|c|<1,
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c
n=0
∴
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(b
1+b
2+b
3+…+b
n)=
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(b
1+b
3+b
5+…)+
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(b
2+b
4+…)
=
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+
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≤3.
解得c≤
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或c>1.
∵0<|c|<1,∴0<c≤
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或-1<c<0.
故c的取值范圍是(-1,0)∪(0,
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].
點評:本題綜合性強,涉及的知識面廣.本題的關(guān)鍵在于根據(jù)韋達(dá)定理求出數(shù)列{a
n}和{b
n}的特點,進(jìn)行數(shù)列分組求和,將題設(shè)中的極限不等式轉(zhuǎn)化為關(guān)于c的不等式,顯然“橋梁”應(yīng)是一元二次方程根與系數(shù)的關(guān)系.