高中數(shù)學(xué)數(shù)列練習(xí)題集與解析

數(shù)列練習(xí)題一.選擇題(共16小題)1.數(shù)列{an}得首項(xiàng)為3,{bn}為等差數(shù)列且bn=an+1﹣an(n∈N*),若b3=﹣2,b10=12,則a8=()A.0B.3C.8D.112.在數(shù)列{an}中,a1=2,an+1=an+ln(1+),則an=()A.2+lnnB.2+(n﹣1)lnnC.2+nlnnD.1+n+lnn3.已知數(shù)列{an}得前n項(xiàng)與Sn=n2﹣9n,第k項(xiàng)滿足5＜ak＜8,則k等于()A.9B.8C.7D.64.已知數(shù)列{an}得前n項(xiàng)與為Sn,a1=1,Sn=2an+1,則Sn=()A.2n﹣1B.C.D.5.已知數(shù)列{an}滿足a1=1,且,且n∈N*),則數(shù)列{an}得通項(xiàng)公式為()A.an=B.an=C.an=n+2D.an=(n+2)3n6.已知數(shù)列{an}中,a1=2,an+1﹣2an=0,bn=log2an,那么數(shù)列{bn}得前10項(xiàng)與等于()A.130B.120C.55D.507.在數(shù)列中,若,則該數(shù)列得通項(xiàng)()A.B.C.D.8.在數(shù)列{an}中,若a1=1,a2=,=+(n∈N*),則該數(shù)列得通項(xiàng)公式為()A.an=B.an=C.an=D.an=9.已知數(shù)列{an}滿足an+1=an﹣an﹣1(n≥2),a1=1,a2=3,記Sn=a1+a2+…+an,則下列結(jié)論正確得就是()A.a100=﹣1,S100=5B.a100=﹣3,S100=5C.a100=﹣3,S100=2D.a100=﹣1,S100=210.已知數(shù)列{an}中,a1=3,an+1=2an+1,則a3=()A.3B.7C.15D.1811.已知數(shù)列{an},滿足an+1=,若a1=,則a2014=()A.B.2C.﹣1D.112.已知數(shù)列中,,,,則=()A.B.C.D.13.已知數(shù)列中,;數(shù)列中,。當(dāng)時(shí),,,求,、()14.已知:數(shù)列{an}滿足a1=16,an+1﹣an=2n,則得最小值為()A.8B.7C.6D.515.已知數(shù)列{an}中,a1=2,nan+1=(n+1)an+2,n∈N+,則a11=()A.36B.38C.40D.4216.已知數(shù)列{an}得前n項(xiàng)與為Sn,a1=1,當(dāng)n≥2時(shí),an+2Sn﹣1=n,則S2015得值為()A.2015B.2013C.1008D.1007二.填空題(共8小題)17.已知無窮數(shù)列{an}前n項(xiàng)與,則數(shù)列{an}得各項(xiàng)與為18.若數(shù)列{an}中,a1=3,且an+1=an2(n∈N*),則數(shù)列得通項(xiàng)an=.19.數(shù)列{an}滿足a1=3,﹣=5(n∈N+),則an=.20.已知數(shù)列{an}得前n項(xiàng)與Sn=n2﹣2n+2,則數(shù)列得通項(xiàng)an=.21.已知數(shù)列{an}中,,則a16=.22.已知數(shù)列{an}得通項(xiàng)公式an=,若它得前n項(xiàng)與為10,則項(xiàng)數(shù)n為.23.數(shù)列{an}滿足an+1+(﹣1)nan=2n﹣1,則{an}得前60項(xiàng)與為.24.已知數(shù)列{an},{bn}滿足a1=,an+bn=1,bn+1=(n∈N*),則b2012=.三.解答題(共6小題)25.設(shè)數(shù)列{an}得前n項(xiàng)與為Sn,n∈N*.已知a1=1,a2=,a3=,且當(dāng)a≥2時(shí),4Sn+2+5Sn=8Sn+1+Sn﹣1.(1)求a4得值;(2)證明:{an+1﹣an}為等比數(shù)列;(3)求數(shù)列{an}得通項(xiàng)公式.26.數(shù)列{an}滿足a1=1,a2=2,an+2=2an+1﹣an+2.(Ⅰ)設(shè)bn=an+1﹣an,證明{bn}就是等差數(shù)列;(Ⅱ)求{an}得通項(xiàng)公式.27.在數(shù)列{an}中,a1=1,an+1=(1+)an+.(1)設(shè)bn=,求數(shù)列{bn}得通項(xiàng)公式;(2)求數(shù)列{an}得前n項(xiàng)與Sn.28.(2015?瓊海校級(jí)模擬)已知正項(xiàng)數(shù)列滿足4Sn=(an+1)2.(1)求數(shù)列{an}得通項(xiàng)公式;(2)設(shè)bn=,求數(shù)列{bn}得前n項(xiàng)與Tn.29.已知{an}就是等差數(shù)列,公差為d,首項(xiàng)a1=3,前n項(xiàng)與為Sn.令,{cn}得前20項(xiàng)與T20=330.數(shù)列{bn}滿足bn=2(a﹣2)dn﹣2+2n﹣1,a∈R.(Ⅰ)求數(shù)列{an}得通項(xiàng)公式;(Ⅱ)若bn+1≤bn,n∈N*,求a得取值范圍.30.已知數(shù)列{an}中,a1=3,前n與Sn=(n+1)(an+1)﹣1.①求證:數(shù)列{an}就是等差數(shù)列②求數(shù)列{an}得通項(xiàng)公式③設(shè)數(shù)列{}得前n項(xiàng)與為Tn,就是否存在實(shí)數(shù)M,使得Tn≤M對(duì)一切正整數(shù)n都成立？若存在,求M得最小值,若不存在,試說明理由.2015年08月23日1384186492得高中數(shù)學(xué)組卷參考答案與試題解析一.選擇題(共16小題)1.(2014?湖北模擬)數(shù)列{an}得首項(xiàng)為3,{bn}為等差數(shù)列且bn=an+1﹣an(n∈N*),若b3=﹣2,b10=12,則a8=()A.0B.3C.8D.11(累加)考點(diǎn):數(shù)列遞推式.專題:計(jì)算題.分析:先利用等差數(shù)列得通項(xiàng)公式分別表示出b3與b10,聯(lián)立方程求得b1與d,進(jìn)而利用疊加法求得b1+b2+…+bn=an+1﹣a1,最后利用等差數(shù)列得求與公式求得答案.解答:解:依題意可知求得b1=﹣6,d=2∵bn=an+1﹣an,∴b1+b2+…+bn=an+1﹣a1,∴a8=b1+b2+…+b7+3=+3=3故選B.點(diǎn)評(píng):本題主要考查了數(shù)列得遞推式.考查了考生對(duì)數(shù)列基礎(chǔ)知識(shí)得熟練掌握.2.(2008?江西)在數(shù)列{an}中,a1=2,an+1=an+ln(1+),則an=()A.2+lnnB.2+(n﹣1)lnnC.2+nlnnD.1+n+lnn(累加)考點(diǎn):數(shù)列得概念及簡單表示法.專題:點(diǎn)列、遞歸數(shù)列與數(shù)學(xué)歸納法.分析:把遞推式整理,先整理對(duì)數(shù)得真數(shù),通分變成,用迭代法整理出結(jié)果,約分后選出正確選項(xiàng).解答:解:∵,,…∴=故選:A.點(diǎn)評(píng):數(shù)列得通項(xiàng)an或前n項(xiàng)與Sn中得n通常就是對(duì)任意n∈N成立,因此可將其中得n換成n+1或n﹣1等,這種辦法通常稱迭代或遞推.解答本題需了解數(shù)列得遞推公式,明確遞推公式與通項(xiàng)公式得異同;會(huì)根據(jù)數(shù)列得遞推公式寫出數(shù)列得前幾項(xiàng).3.(2007?廣東)已知數(shù)列{an}得前n項(xiàng)與Sn=n2﹣9n,第k項(xiàng)滿足5＜ak＜8,則k等于()A.9B.8C.7D.6考點(diǎn):數(shù)列遞推式.專題:計(jì)算題.分析:先利用公式an=求出an,再由第k項(xiàng)滿足5＜ak＜8,求出k.解答:解:an==∵n=1時(shí)適合an=2n﹣10,∴an=2n﹣10.∵5＜ak＜8,∴5＜2k﹣10＜8,∴＜k＜9,又∵k∈N+,∴k=8,故選B.點(diǎn)評(píng):本題考查數(shù)列得通項(xiàng)公式得求法,解題時(shí)要注意公式an=得合理運(yùn)用.4.(2015?房山區(qū)一模)已知數(shù)列{an}得前n項(xiàng)與為Sn,a1=1,Sn=2an+1,則Sn=()A.2n﹣1B.C.D.考點(diǎn):數(shù)列遞推式;等差數(shù)列得通項(xiàng)公式;等差數(shù)列得前n項(xiàng)與.專題:計(jì)算題.分析:直接利用已知條件求出a2,通過Sn=2an+1,推出數(shù)列就是等比數(shù)列,然后求出Sn.解答:解:因?yàn)閿?shù)列{an}得前n項(xiàng)與為Sn,a1=1,Sn=2an+1,a2=所以Sn﹣1=2an,n≥2,可得an=2an+1﹣2an,即:,所以數(shù)列{an}從第2項(xiàng)起,就是等比數(shù)列,所以Sn=1+=,n∈N+.故選:B.點(diǎn)評(píng):本題考查數(shù)列得遞推關(guān)系式得應(yīng)用,前n項(xiàng)與得求法,考查計(jì)算能力.5.(2015?衡水四模)已知數(shù)列{an}滿足a1=1,且,且n∈N*),則數(shù)列{an}得通項(xiàng)公式為()A.an=B.an=C.an=n+2D.an=(n+2)3n考點(diǎn):數(shù)列遞推式.分析:由題意及足a1=1,且,且n∈N*),則構(gòu)造新得等差數(shù)列進(jìn)而求解.解答:解:因?yàn)?且n∈N*)?,即,則數(shù)列{bn}為首項(xiàng),公差為1得等差數(shù)列,所以bn=b1+(n﹣1)×1=3+n﹣1=n+2,所以,故答案為:B點(diǎn)評(píng):此題考查了構(gòu)造新得等差數(shù)列,等差數(shù)列得通項(xiàng)公式.6.(2015?江西一模)已知數(shù)列{an}中,a1=2,an+1﹣2an=0,bn=log2an,那么數(shù)列{bn}得前10項(xiàng)與等于()A.130B.120C.55D.50考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:等差數(shù)列與等比數(shù)列.分析:由題意可得,可得數(shù)列{an}就是以2為首項(xiàng),2為公比得等比數(shù)列,利用等比數(shù)列得通項(xiàng)公式即可得到an,利用對(duì)數(shù)得運(yùn)算法則即可得到bn,再利用等差數(shù)列得前n項(xiàng)公式即可得出.解答:解:在數(shù)列{an}中,a1=2,an+1﹣2an=0,即,∴數(shù)列{an}就是以2為首項(xiàng),2為公比得等比數(shù)列,∴=2n.∴=n.∴數(shù)列{bn}得前10項(xiàng)與=1+2+…+10==55.故選C.點(diǎn)評(píng):熟練掌握等比數(shù)列得定義、等比數(shù)列得通項(xiàng)公式、對(duì)數(shù)得運(yùn)算法則、等差數(shù)列得前n項(xiàng)公式即可得出.7.在數(shù)列中,若,則該數(shù)列得通項(xiàng)()A.B.C.D.8.(2015?遵義校級(jí)二模)在數(shù)列{an}中,若a1=1,a2=,=+(n∈N*),則該數(shù)列得通項(xiàng)公式為()A.an=B.an=C.an=D.an=考點(diǎn):數(shù)列遞推式.專題:計(jì)算題;等差數(shù)列與等比數(shù)列.分析:由=+,確定數(shù)列{}就是等差數(shù)列,即可求出數(shù)列得通項(xiàng)公式.解答:解:∵=+,∴數(shù)列{}就是等差數(shù)列,∵a1=1,a2=,∴=n,∴an=,故選:A.點(diǎn)評(píng):本題考查數(shù)列遞推式,考查數(shù)列得通項(xiàng)公式,確定數(shù)列{}就是等差數(shù)列就是關(guān)鍵.9.(2015?錦州一模)已知數(shù)列{an}滿足an+1=an﹣an﹣1(n≥2),a1=1,a2=3,記Sn=a1+a2+…+an,則下列結(jié)論正確得就是()A.a100=﹣1,S100=5B.a100=﹣3,S100=5C.a100=﹣3,S100=2D.a100=﹣1,S100=2考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:等差數(shù)列與等比數(shù)列.分析:由an+1=an﹣an﹣1(n≥2)可推得該數(shù)列得周期為6,易求該數(shù)列得前6項(xiàng),由此可求得答案.解答:解:由an+1=an﹣an﹣1(n≥2),得an+6=an+5﹣an+4=an+4﹣an+3﹣an+4=﹣an+3=﹣(an+2﹣an+1)=﹣(an+1﹣an﹣an+1)=an,所以6為數(shù)列{an}得周期,又a3=a2﹣a1=3﹣1=2,a4=a3﹣a2=2﹣3=﹣1,a5=a4﹣a3=﹣1﹣2=﹣3,a6=a5﹣a4=﹣3﹣(﹣1)=﹣2,所以a100=a96+4=a4=﹣1,S100=16(a1+a2+a3+a4+a5+a6)+a1+a2+a3+a4=16×0+1+3+2﹣1=5,故選A.點(diǎn)評(píng):本題考查數(shù)列遞推式、數(shù)列求與,考查學(xué)生分析解決問題得能力.10.(2015春?滄州期末)已知數(shù)列{an}中,a1=3,an+1=2an+1,則a3=()A.3B.7C.15D.18考點(diǎn):數(shù)列得概念及簡單表示法.專題:點(diǎn)列、遞歸數(shù)列與數(shù)學(xué)歸納法.分析:根據(jù)數(shù)列得遞推關(guān)系即可得到結(jié)論.解答:解:∵a1=3,an+1=2an+1,∴a2=2a1+1=2×3+1=7,a3=2a2+1=2×7+1=15,故選:C.點(diǎn)評(píng):本題主要考查數(shù)列得計(jì)算,利用數(shù)列得遞推公式就是解決本題得關(guān)鍵,比較基礎(chǔ).11.(2015春?巴中校級(jí)期末)已知數(shù)列{an},滿足an+1=,若a1=,則a2014=()A.B.2C.﹣1D.1考點(diǎn):數(shù)列遞推式.專題:等差數(shù)列與等比數(shù)列.分析:由已知條件,分別令n=1,2,3,4,利用遞推思想依次求出數(shù)列得前5項(xiàng),由此得到數(shù)列{an}就是周期為3得周期數(shù)列,由此能求出a2014.解答:解:∵數(shù)列{an},滿足an+1=,a1=,∴a2==2,a3==﹣1,a4==,,∴數(shù)列{an}就是周期為3得周期數(shù)列,∵2014÷3=671…1,∴a2014=a1=.故選:A.點(diǎn)評(píng):本題考查數(shù)列得第2014項(xiàng)得求法,就是中檔題,解題時(shí)要認(rèn)真審題,注意遞推思想得合理運(yùn)用.12.已知數(shù)列中,,,,則=()A.B.C.D.13.已知數(shù)列中,;數(shù)列中,。當(dāng)時(shí),,,求,、()A.C.B.解:因所以即…………(1)又因?yàn)樗浴?、即……?2)由(1)、(2)得:,14.(2014?通州區(qū)二模)已知:數(shù)列{an}滿足a1=16,an+1﹣an=2n,則得最小值為()A.8B.7C.6D.5考點(diǎn):數(shù)列遞推式.專題:計(jì)算題;壓軸題.分析:a2﹣a1=2,a3﹣a2=4,…,an+1﹣an=2n,這n個(gè)式子相加,就有an+1=16+n(n+1),故,由此能求出得最小值.解答:解:a2﹣a1=2,a3﹣a2=4,…an+1﹣an=2n,這n個(gè)式子相加,就有an+1=16+n(n+1),即an=n(n﹣1)+16=n2﹣n+16,∴,用均值不等式,知道它在n=4得時(shí)候取最小值7.故選B.點(diǎn)評(píng):本題考查數(shù)更列得性質(zhì)與應(yīng)用,解題時(shí)要注意遞推公式得靈活運(yùn)用.15.(2014?中山模擬)已知數(shù)列{an}中,a1=2,nan+1=(n+1)an+2,n∈N+,則a11=()A.36B.38C.40D.42考點(diǎn):數(shù)列遞推式.專題:綜合題;等差數(shù)列與等比數(shù)列.分析:在等式得兩邊同時(shí)除以n(n+1),得﹣=2(﹣),然后利用累加法求數(shù)列得通項(xiàng)公式即可.解答:解:因?yàn)閚an+1=(n+1)an+2(n∈N*),所以在等式得兩邊同時(shí)除以n(n+1),得﹣=2(﹣),所以=+2[(﹣)+(﹣)+…+(1﹣)]=所以a11=42故選D.點(diǎn)評(píng):本題主要考查利用累加法求數(shù)列得通項(xiàng)公式,以及利用裂項(xiàng)法求數(shù)列得與,要使熟練掌握這些變形技巧.16.(2015?綏化一模)已知數(shù)列{an}得前n項(xiàng)與為Sn,a1=1,當(dāng)n≥2時(shí),an+2Sn﹣1=n,則S2015得值為()A.2015B.2013C.1008D.1007考點(diǎn):數(shù)列遞推式.專題:點(diǎn)列、遞歸數(shù)列與數(shù)學(xué)歸納法.分析:根據(jù)an+2Sn﹣1=n得到遞推關(guān)系an+1+an=1,n≥2,從而得到當(dāng)n就是奇數(shù)時(shí),an=1,n就是偶數(shù)時(shí),an=0,即可得到結(jié)論.解答:解:∵當(dāng)n≥2時(shí),an+2Sn﹣1=n,∴an+1+2Sn=n+1,兩式相減得:an+1+2Sn﹣(an+2Sn﹣1)=n+1﹣n,即an+1+an=1,n≥2,當(dāng)n=2時(shí),a2+2a1=2,解得a2=2﹣2a1=0,滿足an+1+an=1,則當(dāng)n就是奇數(shù)時(shí),an=1,當(dāng)n就是偶數(shù)時(shí),an=0,則S2015=1008,故選:C點(diǎn)評(píng):本題主要考查數(shù)列與得計(jì)算,根據(jù)數(shù)列得遞推關(guān)系求出數(shù)列項(xiàng)得特點(diǎn)就是解決本題得關(guān)鍵.二.填空題(共8小題)17.(2008?上海)已知無窮數(shù)列{an}前n項(xiàng)與,則數(shù)列{an}得各項(xiàng)與為﹣1考點(diǎn):數(shù)列遞推式;極限及其運(yùn)算.專題:計(jì)算題.分析:若想求數(shù)列得前N項(xiàng)與,則應(yīng)先求數(shù)列得通項(xiàng)公式an,由已知條件,結(jié)合an=Sn﹣Sn﹣1可得遞推公式,因?yàn)榫褪乔鬅o窮遞縮等比數(shù)列得所有項(xiàng)得與,故由公式S=即得解答:解:由可得:(n≥2),兩式相減得并化簡:(n≥2),又,所以無窮數(shù)列{an}就是等比數(shù)列,且公比為﹣,即無窮數(shù)列{an}為遞縮等比數(shù)列,所以所有項(xiàng)得與S=故答案就是﹣1點(diǎn)評(píng):本題主要借助數(shù)列前N項(xiàng)與與項(xiàng)得關(guān)系,考查了數(shù)列得遞推公式與無窮遞縮等比數(shù)列所有項(xiàng)與公式,并檢測了學(xué)生對(duì)求極限知識(shí)得掌握,屬于一個(gè)比較綜合得問題.18.(2002?上海)若數(shù)列{an}中,a1=3,且an+1=an2(n∈N*),則數(shù)列得通項(xiàng)an=.考點(diǎn):數(shù)列遞推式.專題:計(jì)算題;壓軸題.分析:由遞推公式an+1=an2多次運(yùn)用迭代可求出數(shù)列an=an﹣12=an﹣24=…=a12n﹣1解答:解:因?yàn)閍1=3多次運(yùn)用迭代,可得an=an﹣12=an﹣24=…=a12n﹣1=32n﹣1,故答案為:點(diǎn)評(píng):本題主要考查利用迭代法求數(shù)列得通項(xiàng)公式,迭代中要注意規(guī)律,靈活運(yùn)用公式,熟練變形就是解題得關(guān)鍵19.(2015?張掖二模)數(shù)列{an}滿足a1=3,﹣=5(n∈N+),則an=.考點(diǎn):數(shù)列遞推式;等差數(shù)列得通項(xiàng)公式.專題:計(jì)算題.分析:根據(jù)所給得數(shù)列得遞推式,瞧出數(shù)列就是一個(gè)等差數(shù)列,根據(jù)所給得原來數(shù)列得首項(xiàng)瞧出等差數(shù)列得首項(xiàng),根據(jù)等差數(shù)列得通項(xiàng)公式寫出數(shù)列,進(jìn)一步得到結(jié)果.解答:解:∵根據(jù)所給得數(shù)列得遞推式∴數(shù)列{}就是一個(gè)公差就是5得等差數(shù)列,∵a1=3,∴=,∴數(shù)列得通項(xiàng)就是∴故答案為:點(diǎn)評(píng):本題瞧出數(shù)列得遞推式與數(shù)列得通項(xiàng)公式,本題解題得關(guān)鍵就是確定數(shù)列就是一個(gè)等差數(shù)列,利用等差數(shù)列得通項(xiàng)公式寫出通項(xiàng),本題就是一個(gè)中檔題目.20.(2015?歷下區(qū)校級(jí)四模)已知數(shù)列{an}得前n項(xiàng)與Sn=n2﹣2n+2,則數(shù)列得通項(xiàng)an=.考點(diǎn):數(shù)列遞推式.專題:計(jì)算題.分析:由已知中數(shù)列{an}得前n項(xiàng)與Sn=n2﹣2n+2,我們可以根據(jù)an=求出數(shù)列得通項(xiàng)公式,但最后要驗(yàn)證n=1時(shí),就是否滿足n≥2時(shí)所得得式子,如果不滿足,則寫成分段函數(shù)得形式.解答:解:∵Sn=n2﹣2n+2,∴當(dāng)n≥2時(shí),an=Sn﹣Sn﹣1=(n2﹣2n+2)﹣[(n﹣1)2﹣2(n﹣1)+2]=2n﹣3又∵當(dāng)n=1時(shí)a1=S1=1≠2×1﹣3故an=故答案為:點(diǎn)評(píng):本題考查得知識(shí)點(diǎn)就是由前n項(xiàng)與公式,求數(shù)列得通項(xiàng)公式,其中掌握an=,及解答此類問題得步驟就是關(guān)鍵.21.(2015春?邢臺(tái)校級(jí)月考)已知數(shù)列{an}中,,則a16=.考點(diǎn):數(shù)列遞推式.專題:計(jì)算題.分析:由,可分別求a2,a3,a4,從而可得數(shù)列得周期,可求解答:解:∵,則=﹣1=2=∴數(shù)列{an}就是以3為周期得數(shù)列∴a16=a1=故答案為:點(diǎn)評(píng):本題主要考查了利用數(shù)列得遞推公式求解數(shù)列得項(xiàng),其中尋求數(shù)列得項(xiàng)得規(guī)律,找出數(shù)列得周期就是求解得關(guān)鍵22.(2014春?庫爾勒市校級(jí)期末)已知數(shù)列{an}得通項(xiàng)公式an=,若它得前n項(xiàng)與為10,則項(xiàng)數(shù)n為120.考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:計(jì)算題.分析:由題意知an=,所以Sn=(﹣)+(﹣)+()=﹣1,再由﹣1=10,可得n=120.解答:解:∵an==∴Sn=(﹣)+(﹣)+()=﹣1∴﹣1=10,解得n=120答案:120點(diǎn)評(píng):本題考查數(shù)列得性質(zhì)與應(yīng)用,解題時(shí)要認(rèn)真審題,仔細(xì)解答.23.(2012?黑龍江)數(shù)列{an}滿足an+1+(﹣1)nan=2n﹣1,則{an}得前60項(xiàng)與為1830.考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:計(jì)算題;壓軸題.分析:令bn+1=a4n+1+a4n+2+a4n+3+a4n+4,則bn+1=a4n+1+a4n+2+a4n+3+a4n+4=a4n﹣3+a4n﹣2+a4n﹣2+a4n+16=bn+16可得數(shù)列{bn}就是以16為公差得等差數(shù)列,而{an}得前60項(xiàng)與為即為數(shù)列{bn}得前15項(xiàng)與,由等差數(shù)列得求與公式可求解答:解:∵,∴令bn+1=a4n+1+a4n+2+a4n+3+a4n+4,a4n+1+a4n+3=(a4n+3+a4n+2)﹣(a4n+2﹣a4n+1)=2,a4n+2+a4n+4=(a4n+4﹣a4n+3)+(a4n+3+a4n+2)=16n+8,則bn+1=a4n+1+a4n+2+a4n+3+a4n+4=a4n﹣3+a4n﹣2+a4n﹣1+a4n+16=bn+16∴數(shù)列{bn}就是以16為公差得等差數(shù)列,{an}得前60項(xiàng)與為即為數(shù)列{bn}得前15項(xiàng)與∵b1=a1+a2+a3+a4=10∴=1830點(diǎn)評(píng):本題主要考查了由數(shù)列得遞推公式求解數(shù)列得與,等差數(shù)列得求與公式得應(yīng)用,解題得關(guān)鍵就是通過構(gòu)造等差數(shù)列24.(2012?浙江模擬)已知數(shù)列{an},{bn}滿足a1=,an+bn=1,bn+1=(n∈N*),則b2012=.;考點(diǎn):數(shù)列遞推式.專題:綜合題.分析:根據(jù)數(shù)列遞推式,判斷{}就是以﹣2為首項(xiàng),﹣1為公差得等差數(shù)列,即可求得,故可求結(jié)論.解答:解:∵an+bn=1,bn+1=∴bn+1==∴bn+1﹣1=∴﹣=﹣1∵=﹣2∴{}就是以﹣2為首項(xiàng),﹣1為公差得等差數(shù)列∴∴∴b2012=故答案為:點(diǎn)評(píng):本題考查數(shù)列遞推式,解題得關(guān)鍵就是判定{}就是以﹣2為首項(xiàng),﹣1為公差得等差數(shù)列,屬于中檔題.三.解答題(共6小題)25.(2015?廣東)設(shè)數(shù)列{an}得前n項(xiàng)與為Sn,n∈N*.已知a1=1,a2=,a3=,且當(dāng)a≥2時(shí),4Sn+2+5Sn=8Sn+1+Sn﹣1.(1)求a4得值;(2)證明:{an+1﹣an}為等比數(shù)列;(3)求數(shù)列{an}得通項(xiàng)公式.考點(diǎn):數(shù)列遞推式.專題:等差數(shù)列與等比數(shù)列.分析:(1)直接在數(shù)列遞推式中取n=2,求得;(2)由4Sn+2+5Sn=8Sn+1+Sn﹣1(n≥2),變形得到4an+2+an=4an+1(n≥2),進(jìn)一步得到,由此可得數(shù)列{}就是以為首項(xiàng),公比為得等比數(shù)列;(3)由{}就是以為首項(xiàng),公比為得等比數(shù)列,可得.進(jìn)一步得到,說明{}就是以為首項(xiàng),4為公差得等差數(shù)列,由此可得數(shù)列{an}得通項(xiàng)公式.解答:(1)解:當(dāng)n=2時(shí),4S4+5S2=8S3+S1,即,解得:;(2)證明:∵4Sn+2+5Sn=8Sn+1+Sn﹣1(n≥2),∴4Sn+2﹣4Sn+1+Sn﹣Sn﹣1=4Sn+1﹣4Sn(n≥2),即4an+2+an=4an+1(n≥2),∵,∴4an+2+an=4an+1.∵=.∴數(shù)列{}就是以為首項(xiàng),公比為得等比數(shù)列;(3)解:由(2)知,{}就是以為首項(xiàng),公比為得等比數(shù)列,∴.即,∴{}就是以為首項(xiàng),4為公差得等差數(shù)列,∴,即,∴數(shù)列{an}得通項(xiàng)公式就是.點(diǎn)評(píng):本題考查了數(shù)列遞推式,考查了等比關(guān)系得確定,考查了等比數(shù)列得通項(xiàng)公式,關(guān)鍵就是靈活變形能力,就是中檔題.26.(2014?廣西)數(shù)列{an}滿足a1=1,a2=2,an+2=2an+1﹣an+2.(Ⅰ)設(shè)bn=an+1﹣an,證明{bn}就是等差數(shù)列;(Ⅱ)求{an}得通項(xiàng)公式.考點(diǎn):數(shù)列遞推式;等差數(shù)列得通項(xiàng)公式;等差關(guān)系得確定.專題:等差數(shù)列與等比數(shù)列.分析:(Ⅰ)將an+2=2an+1﹣an+2變形為:an+2﹣an+1=an+1﹣an+2,再由條件得bn+1=bn+2,根據(jù)條件求出b1,由等差數(shù)列得定義證明{bn}就是等差數(shù)列;(Ⅱ)由(Ⅰ)與等差數(shù)列得通項(xiàng)公式求出bn,代入bn=an+1﹣an并令n從1開始取值,依次得(n﹣1)個(gè)式子,然后相加,利用等差數(shù)列得前n項(xiàng)與公式求出{an}得通項(xiàng)公式an.解答:解:(Ⅰ)由an+2=2an+1﹣an+2得,an+2﹣an+1=an+1﹣an+2,由bn=an+1﹣an得,bn+1=bn+2,即bn+1﹣bn=2,又b1=a2﹣a1=1,所以{bn}就是首項(xiàng)為1,公差為2得等差數(shù)列.(Ⅱ)由(Ⅰ)得,bn=1+2(n﹣1)=2n﹣1,由bn=an+1﹣an得,an+1﹣an=2n﹣1,則a2﹣a1=1,a3﹣a2=3,a4﹣a3=5,…,an﹣an﹣1=2(n﹣1)﹣1,所以,an﹣a1=1+3+5+…+2(n﹣1)﹣1==(n﹣1)2,又a1=1,所以{an}得通項(xiàng)公式an=(n﹣1)2+1=n2﹣2n+2.點(diǎn)評(píng):本題考查了等差數(shù)列得定義、通項(xiàng)公式、前n項(xiàng)與公式,及累加法求數(shù)列得通項(xiàng)公式與轉(zhuǎn)化思想,屬于中檔題.27.(2012?碑林區(qū)校級(jí)模擬)在數(shù)列{an}中,a1=1,an+1=(1+)an+.(1)設(shè)bn=,求數(shù)列{bn}得通項(xiàng)公式;(2)求數(shù)列{an}得前n項(xiàng)與Sn.考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:計(jì)算題;綜合題.分析:(1)由已知得=+,即bn+1=bn+,由此能夠推導(dǎo)出所求得通項(xiàng)公式.(2)由題設(shè)知an=2n﹣,故Sn=(2+4+…+2n)﹣(1++++…+),設(shè)Tn=1++++…+,由錯(cuò)位相減法能求出Tn=4﹣.從而導(dǎo)出數(shù)列{an}得前n項(xiàng)與Sn.解答:解:(1)由已知得b1=a1=1,且=+,即bn+1=bn+,從而b2=b1+,b3=b2+,bn=bn﹣1+(n≥2).于就是bn=b1+++…+=2﹣(n≥2).又b1=1,故所求得通項(xiàng)公式為bn=2﹣.(2)由(1)知an=2n﹣,故Sn=(2+4+…+2n)﹣(1++++…+),設(shè)Tn=1++++…+,①Tn=+++…++,②①﹣②得,Tn=1++++…+﹣=﹣=2﹣﹣,∴Tn=4﹣.∴Sn=n(n+1)+﹣4.點(diǎn)評(píng):本題考查數(shù)列得通項(xiàng)公式與前n項(xiàng)與得求法,解題時(shí)要注意錯(cuò)位相減法得合理運(yùn)用.28.(2015?瓊海校級(jí)模擬)已知正項(xiàng)數(shù)列滿足4Sn=(an+1)2.(1)求數(shù)列{an}得通項(xiàng)公式;(2)設(shè)bn=,求數(shù)列{bn}得前n項(xiàng)與Tn.考點(diǎn):數(shù)列遞推式;數(shù)列得求與.專題:計(jì)算題;等差數(shù)列與等比數(shù)列.分析:(Ⅰ)由4Sn=(an+1)2.可知當(dāng)n≥2時(shí),4Sn﹣1=(an﹣1+1)2,兩式相減,結(jié)合等差數(shù)列得通項(xiàng)公式可求(Ⅱ)由(1)知=,利用裂項(xiàng)求與即可求解解答:解:(Ⅰ)∵4Sn=(an+1)2.∴當(dāng)n≥2時(shí),4Sn﹣1=(an﹣1+1)2.兩式相減可得,4(sn﹣sn﹣1)=即4an=整理得an﹣an﹣1=2…(4分)又a1=1∴an=1+2(n﹣1)=2n﹣1…(6分)(Ⅱ)由(1)知=…(8分)所以=…(12分)點(diǎn)評(píng):本題主要考查了利用數(shù)列得遞推公式求解數(shù)列得通項(xiàng)公式及等差數(shù)列得通項(xiàng)公式、數(shù)列得裂項(xiàng)求與方法得應(yīng)用29.(2015?揭陽校級(jí)三模)已知{an}就是等差數(shù)列,公差為d,首項(xiàng)a1=3,前n項(xiàng)與為Sn.令,{cn}得前20項(xiàng)與T20=330.數(shù)列{bn}滿足bn=2(a﹣2)dn﹣2+2n﹣1,a∈R.(Ⅰ)求數(shù)列{an}得通項(xiàng)公式;(Ⅱ)若bn+1≤bn,n∈N*,求a得取