已知等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,等比數(shù)列{bn}的前n項(xiàng)和為T(mén)n,且{an}、{bn}滿足條件:S4=4a3-2,Tn=2bn-2.
(Ⅰ)求公差d的值;
(Ⅱ)若對(duì)任意的n∈N*,都有Sn≥S5成立,求a1的取直范圍;
(Ⅲ)若a1=-4,令cn=anbn,求數(shù)列{cn}的前n項(xiàng)和Vn.
【答案】
分析:(I)利用等差數(shù)列的通項(xiàng)公式和前n項(xiàng)和公式即可解出;
(II)利用等差前n項(xiàng)和公式化為(n-5)(2a
1+n+4)≥0.由于對(duì)任意的n∈N
*,都有S
n≥S
5成立,可得
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且
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,解出即可.
(III)利用等差數(shù)列的通項(xiàng)公式即可得出a
n.利用n≥2時(shí),b
n=T
n-T
n-1,n=1時(shí)b
1=T
1,及等比數(shù)列的通項(xiàng)公式即可得到b
n.利用“錯(cuò)位相減法”即可得到V
n.
解答:解:(I)設(shè)等比數(shù)列{b
n}的公比為q,由S
4=4a
3-2,得
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,化為6d=8d-2,解得d=1.即公差d=1.
(II)由S
n≥S
5成立,得到
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,化為(n-5)(2a
1+n+4)≥0.
由于對(duì)任意的n∈N
*,都有S
n≥S
5成立,∴
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且
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解得
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.
∴
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.
(III)①當(dāng)a
1=-4時(shí),a
n=-4+(n-1)×1=n-5;
②當(dāng)n=1時(shí),b
1=T
1=2b
1-2,解得b
1=2;
當(dāng)n≥2時(shí),b
n=T
n-T
n-1=2b
n-2-(2b
n-1-2)=2b
n-2b
n-1,化為b
n=2b
n-1.
∴數(shù)列{b
n}是以2為首項(xiàng),2為公比的等比數(shù)列,∴
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.
∴
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.
∴
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+0+2
6+2×2
7+…+(n-5)•2
n,
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-2
5+2
7+2
8+…+(n-6)•2
n+(n-5)•2
n+1.
兩式相減得-V
n=-8+2
2+2
3+…+2
n+(5-n)•2
n+1=
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,
化為
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.
點(diǎn)評(píng):數(shù)列掌握等差數(shù)列的通項(xiàng)公式和前n項(xiàng)和公式、分類討論的思想方法、利用n≥2時(shí)b
n=T
n-T
n-1及n=1時(shí)b
1=T
1、等比數(shù)列的通項(xiàng)公式、“錯(cuò)位相減法”是解題的關(guān)鍵.