高考總復(fù)習(xí)課程-高考數(shù)學(xué)尖子生拔高課程（文）課后練習(xí)第78講適當(dāng)放縮在數(shù)列中的應(yīng)用（上下）

第7講適當(dāng)放縮在數(shù)列中的應(yīng)用（上）題一：已知，求證：，.題二：已知為正數(shù)，且，試證：對(duì)每一個(gè)，.題三：設(shè)實(shí)數(shù)數(shù)列的前n項(xiàng)和，滿足（I）若成等比數(shù)列，求和；（II）求證：對(duì)題四：已知數(shù)列{an}滿足(n∈N*)，Sn是{an}的前n項(xiàng)的和，a2=1．

（1）求Sn；（2）證明：.題五：設(shè)數(shù)列的前項(xiàng)和為，且．（1）求數(shù)列的通項(xiàng)公式；（2）設(shè)，數(shù)列的前項(xiàng)和為，求證：．題六：已知數(shù)列滿足（1）求；（2）已知存在實(shí)數(shù)，使為公差為的等差數(shù)列，求的值；（3）記，數(shù)列的前項(xiàng)和為，求證：.第8講適當(dāng)放縮在數(shù)列中的應(yīng)用（下）題一：已知數(shù)列是公差不為0的等差數(shù)列，的等比中項(xiàng).（1）求數(shù)列的通項(xiàng)公式；（2）設(shè)題二：已知數(shù)列的首項(xiàng)為，前項(xiàng)和為，且對(duì)任意的，當(dāng)n≥2時(shí)，an總是3Sn－4與2－EQ\f(5,2)Sn的等差中項(xiàng)（Ⅰ）求數(shù)列{an}的通項(xiàng)公式；（Ⅱ）設(shè)，是數(shù)列的前項(xiàng)和，求；（Ⅲ）設(shè)，是數(shù)列的前項(xiàng)和，，，試證明：．題三：設(shè)無窮數(shù)列{an}具有以下性質(zhì)：①a1=1；②當(dāng)EMBEDEquation.3（Ⅰ）請(qǐng)給出一個(gè)具有這種性質(zhì)的無窮數(shù)列，使得不等式EMBEDEquation.3對(duì)于任意的EMBEDEquation.3都成立，并對(duì)你給出的結(jié)果進(jìn)行驗(yàn)證（或證明）；（Ⅱ）若EMBEDEquation.3，其中EMBEDEquation.3，且記數(shù)列{bn}的前n項(xiàng)和Bn，證明：EMBEDEquation.3題四：已知數(shù)列EMBEDEquation.DSMT4滿足EMBEDEquation.DSMT4（Ⅰ）求數(shù)列EMBEDEquation.DSMT4的通項(xiàng)公式；（Ⅱ）若數(shù)列EMBEDEquation.DSMT4滿足EMBEDEquation.DSMT4，證明：EMBEDEquation.DSMT4是等差數(shù)列；（Ⅲ）證明：EMBEDEquation.DSMT4題五：已知數(shù)列EMBEDEquation.DSMT4的首項(xiàng)EMBEDEquation.DSMT4，EMBEDEquation.DSMT4，EMBEDEquation.DSMT4．（Ⅰ）求EMBEDEquation.DSMT4的通項(xiàng)公式；（Ⅱ）證明：對(duì)任意的EMBEDEquation.DSMT4，EMBEDEquation.DSMT4，EMBEDEquation.DSMT4；（Ⅲ）證明：EMBEDEquation.DSMT4．題六：已知數(shù)列{an}滿足a1=5，a2=5，an+1=an+6an－1（n≥2，n∈N*），若數(shù)列是等比數(shù)列.（Ⅰ）求數(shù)列{an}的通項(xiàng)公式；（Ⅱ）求證：當(dāng)k為奇數(shù)時(shí)，；（Ⅲ）求證：

第7講適當(dāng)放縮在數(shù)列中的應(yīng)用（上）題一：證明：EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4，EMBEDEquation.DSMT4.題二：證明：由EMBEDEquation.3得EMBEDEquation.3，又EMBEDEquation.3，故EMBEDEquation.3，而EMBEDEquation.3，令EMBEDEquation.3，則EMBEDEquation.3=EMBEDEquation.3，因?yàn)镋MBEDEquation.3，倒序相加得EMBEDEquation.3=EMBEDEquation.3，而EMBEDEquation.3，則EMBEDEquation.3=EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3，所以EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3，即對(duì)每一個(gè)EMBEDEquation.3，EMBEDEquation.3.題三：見詳解詳解：（I）由題意EMBEDEquation.DSMT4，由S2是等比中項(xiàng)知EMBEDEquation.DSMT4由EMBEDEquation.DSMT4解得EMBEDEquation.DSMT4（II）由題設(shè)條件有EMBEDEquation.DSMT4故EMBEDEquation.DSMT4從而對(duì)EMBEDEquation.DSMT4有EMBEDEquation.DSMT4①因EMBEDEquation.DSMT4，由①得EMBEDEquation.DSMT4要證EMBEDEquation.DSMT4，由①只要證EMBEDEquation.DSMT4即證EMBEDEquation.DSMT4此式明顯成立.因此EMBEDEquation.DSMT4最后證EMBEDEquation.DSMT4若不然EMBEDEquation.DSMT4又因EMBEDEquation.DSMT4矛盾.因此EMBEDEquation.DSMT4題四：EMBEDEquation.DSMT4詳解：（1）由題意EMBEDEquation.DSMT4得EMBEDEquation.DSMT4，

兩式相減得EMBEDEquation.DSMT4即（n1）an+1=nan，

所以（n+1）an+1=nan+2再相加得2nan+1=nan+nan+2即2an+1=an+an+2

所以數(shù)列{an}是等差數(shù)列

∵a1=EMBEDEquation.DSMT4a1∴a1=0，又a2=1，則公差為1，∴an=n1，

所以數(shù)列{an}的前n項(xiàng)的和為EMBEDEquation.DSMT4（2）EMBEDEquation.DSMT4①當(dāng)n=1時(shí)：EMBEDEquation.DSMT4，EMBEDEquation.DSMT4，不等式成立．

②當(dāng)n≥2時(shí)：一方面

∵EMBEDEquation.DSMT4另一方面：EMBEDEquation.DSMT4

∴EMBEDEquation.DSMT4，

綜合兩方面∴EMBEDEquation.DSMT4.于是對(duì)于正整數(shù)n，都有EMBEDEquation.DSMT4題五：EMBEDEquation.3詳解：當(dāng)EMBEDEquation.DSMT4時(shí)，EMBEDEquation.DSMT4．當(dāng)EMBEDEquation.DSMT4時(shí)，EMBEDEquation.3 EMBEDEquation.3 EMBEDEquation.3． ∵EMBEDEquation.3不適合上式, ∴EMBEDEquation.3（2）證明:∵EMBEDEquation.DSMT4． 當(dāng)EMBEDEquation.3時(shí)，EMBEDEquation.DSMT4 當(dāng)EMBEDEquation.DSMT4時(shí)，EMBEDEquation.DSMT4,①EMBEDEquation.DSMT4．②①－②得:EMBEDEquation.DSMT4EMBEDEquation.DSMT4得EMBEDEquation.DSMT4，此式當(dāng)EMBEDEquation.3時(shí)也適合．∴EMBEDEquation.3NEMBEDEquation.3．∵EMBEDEquation.DSMT4，∴EMBEDEquation.DSMT4． 當(dāng)EMBEDEquation.DSMT4時(shí)，EMBEDEquation.DSMT4， ∴EMBEDEquation.DSMT4．∵EMBEDEquation.DSMT4，∴EMBEDEquation.DSMT4． 故EMBEDEquation.DSMT4，即EMBEDEquation.DSMT4．綜上，EMBEDEquation.DSMT4．題六：詳解：（1），由數(shù)列的遞推公式得，，.（2）===.數(shù)列為公差是的等差數(shù)列.由題意，令，得.（3）由（2）知，所以.此時(shí)EMBEDEquation.DSMT4==，=>.第8講適當(dāng)放縮在數(shù)列中的應(yīng)用（下）題一：EMBEDEquation.DSMT4詳解：設(shè)EMBEDEquation.DSMT4解得EMBEDEquation.DSMT4（2）證明：EMBEDEquation.DSMT4EMBEDEquation.DSMT4題二：an＝2EMBEDEquation.DSMT4詳解：（Ⅰ）當(dāng)n≥2時(shí)，2an＝3Sn－4＋2－EQ\f(5,2)Sn，即2(Sn－Sn－1)＝3Sn－4＋2－EQ\f(5,2)Sn，所以Sn＝EQ\f(1,2)Sn－1＋2∴EQ\f(an＋1,an)＝\f(Sn＋1－Sn,Sn－Sn－1)＝\f((\f(1,2)Sn＋2)－(\f(1,2)Sn－1＋2),Sn－Sn－1)＝\f(1,2)(n≥2)又2＋a2＝EQ\f(1,2)×2＋2＝3a2＝1EQ\f(a2,a1)＝\f(1,2)∴數(shù)列{an}是首項(xiàng)為2，公比為EQ\f(1,2)的等比數(shù)列∴an＝2EMBEDEquation.DSMT4(n∈N*)（Ⅱ）由（Ⅰ）知an＝22－n(n∈N*)則Tn＝b1＋b2＋……＋bn＝2×2＋3×1＋4×EQ\f(1,2)＋……＋(n＋1)×2EMBEDEquation.DSMT4∴EQ\f(1,2)Tn＝2×1＋3×EQ\f(1,2)＋……＋n×23－n＋(n＋1)×2EMBEDEquation.DSMT4,作差得：EQ\f(1,2)Tn＝2×2＋1＋EQ\f(1,2)＋EQ\f(1,4)＋……＋23－n－(n＋1)2EMBEDEquation.DSMT4＝6－EQ\f(n＋3,2n－1)∴Tn＝12－EQ\f(n＋3,2n－2)(n∈N*)（Ⅲ）證明：題三：證明：（Ⅰ）令EMBEDEquation.3，則無窮數(shù)列{an}可由a1=1，EMBEDEquation.3給出.顯然，該數(shù)列滿足EMBEDEquation.3，且EMBEDEquation.3（Ⅱ）EMBEDEquation.3EMBEDEquation.3又EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3題四：EMBEDEquation.3詳解：（1）EMBEDEquation.3，EMBEDEquation.3故數(shù)列EMBEDEquation.3是首項(xiàng)為2，公比為2的等比數(shù)列。EMBEDEquation.3，EMBEDEquation.3（2）EMBEDEquation.3，EMBEDEquation.3EMBEDEquation.3①EMBEDEquation.3②②—①得EMBEDEquation.3，即EMBEDEquation.3③EMBEDEquation.3④④—③得EMBEDEquation.3，即EMBEDEquation.DSMT4所以數(shù)列是等差數(shù)列（3）EMBEDEquation.3設(shè)EMBEDEquation.3，則EMBEDEquation.3EMBEDEquation.3EMBEDEquation.3題五：EMBEDEquation.DSMT4詳解：（Ⅰ）EMBEDEquation.DSMT4，EMBEDEquation.DSMT4，EMBEDEquation.DSMT4，又EMBEDEquation.DSMT4，EMBEDEquation.DSMT4是以EMBEDEquation.DSMT4為首項(xiàng)，EMBEDEquation.DSMT4為公比的等比數(shù)列．EMBEDEquation.DSMT4EMBEDEquation.DSMT4，EMBEDEquation.DSMT