20xx年高考數(shù)學(xué)總復(fù)習(xí)數(shù)列的綜合應(yīng)用限時(shí)練習(xí)詳細(xì)解析人教大綱版(存儲(chǔ)版)
【正文】 .恒成立,即關(guān)于n是遞增的.∴an+1＞an.答案:B二、填空題{an}的通項(xiàng)為an＝2n7(n∈N*),則|a1|+|a2|+…+|a15|＝___________.解析:由an＝2n7≤0得n≤,ai≤0(i＝1,2,3),Sn＝a1+a2+…+an＝n2+n7n＝n26n.所以|a1|+|a2|+…+|a15|＝a1a2a3+a4+a5+…+a15＝2S3+S15＝2(9)+135＝153.答案:153,…的前n項(xiàng)和等于________.解析:∵,∴原式＝.答案:{an}的公比為q,前n項(xiàng)和為Sn,且S3,S9,S6成等差數(shù)列,則q等于__________.解析:公比q＝1時(shí),不符合題意,∴q≠1.∴.解得.答案:“*”運(yùn)算:對于n∈N*,滿足以下運(yùn)算性質(zhì):①2*2＝1。(理)(3)設(shè),且數(shù)列{}的前n項(xiàng)和為Sn,求的值.解:(1)由an+12an22an+12an＝0,得(an+1+an)(an+1an2)＝0.因?yàn)閍n＞0,所以an+1an2＝0,an+1an＝2.所以數(shù)列{an}是以a1＝1為首項(xiàng),以2為公差的等差數(shù)列.所以an＝1+(n1)2＝2n1,bn＝n33n2+5an＝n33n2+52n+1＝n33n22n+6.(2)由(1)得bnan＝n33n22n+6(2n1)＝n33n24n+7,當(dāng)n＝1時(shí),b1a1＝134+7＝1＞0b1＞a1。偶數(shù)項(xiàng)組成以a2＝4為首項(xiàng),公比為4的等比數(shù)列.∴.當(dāng)n為偶數(shù)時(shí),奇數(shù)項(xiàng)和偶數(shù)項(xiàng)各有項(xiàng)
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