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《2018屆高三數(shù)學(xué)(理)二輪復(fù)習(xí)專題集訓(xùn):專題四 數(shù)列4.2含解析》由會(huì)員上傳分享,免費(fèi)在線閱讀,更多相關(guān)內(nèi)容在教育資源-天天文庫。
1、2018屆高三數(shù)學(xué)二輪復(fù)習(xí)專題集訓(xùn)A級(jí)1.已知數(shù)列{an}中,a1=a2=1,an+2=則數(shù)列{an}的前20項(xiàng)和為( )A.1121B.1122C.1123D.1124解析: 由題意可知,數(shù)列{a2n}是首項(xiàng)為1,公比為2的等比數(shù)列,數(shù)列{a2n-1}是首項(xiàng)為1,公差為2的等差數(shù)列,故數(shù)列{an}的前20項(xiàng)和為+10×1+×2=1123.選C.答案: C2.若數(shù)列{an}滿足a1=15,且3an+1=3an-2,則使ak·ak+1<0的k值為( )A.22B.21C.24D.23解析: 因?yàn)?an+1=3an-2,所以an+1-an=-,所以數(shù)列{an}
2、是首項(xiàng)為15,公差為-的等差數(shù)列,所以an=15-·(n-1)=-n+,令an=-n+>0,得n<23.5,所以使ak·ak+1<0的k值為23.答案: D3.(2017·廣東省五校協(xié)作體第一次診斷考試)數(shù)列{an}滿足a1=1,且an+1=a1+an+n(n∈N*),則++…+等于( )A.B.C.D.解析: 由a1=1,an+1=a1+an+n可得an+1-an=n+1,利用累加法可得an-a1=,所以an=,所以==2,故++…+=2=2=,選A.答案: A4.(2017·湖北省七市(州)聯(lián)考)在各項(xiàng)都為正數(shù)的數(shù)列{an}中,首項(xiàng)a1=2,且點(diǎn)(a,a
3、)在直線x-9y=0上,則數(shù)列{an}的前n項(xiàng)和Sn等于( )62018屆高三數(shù)學(xué)二輪復(fù)習(xí)專題集訓(xùn)A.3n-1B.C.D.解析: 由點(diǎn)(a,a)在直線x-9y=0上,得a-9a=0,即(an+3an-1)(an-3an-1)=0,又?jǐn)?shù)列{an}各項(xiàng)均為正數(shù),且a1=2,∴an+3an-1>0,∴an-3an-1=0,即=3,∴數(shù)列{an}是首項(xiàng)a1=2,公比q=3的等比數(shù)列,其前n項(xiàng)和Sn===3n-1,故選A.答案: A5.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,若a2=12,a3·a5=4,則下列說法正確的是( )A.{an}是單調(diào)遞減數(shù)列B.{Sn}是
4、單調(diào)遞減數(shù)列C.{a2n}是單調(diào)遞減數(shù)列D.{S2n}是單調(diào)遞減數(shù)列解析: 由于{an}是等比數(shù)列,則a3a5=a=4,又a2=12,則a4>0,a4=2,q2=,當(dāng)q=-時(shí),{an}和{Sn}不具有單調(diào)性,選項(xiàng)A和B錯(cuò)誤;a2n=a2q2n-2=12×n-1單調(diào)遞減,選項(xiàng)C正確;當(dāng)q=-時(shí),{S2n}不具有單調(diào)性,選項(xiàng)D錯(cuò)誤.答案: C6.在數(shù)列{an}中,a1=1,an+2+(-1)nan=1.記Sn是數(shù)列{an}的前n項(xiàng)和,則S100=________.解析: 當(dāng)n=2k時(shí),a2k+2+a2k=1;當(dāng)n=2k-1時(shí),a2k+1=a2k-1+1,所以a2k
5、-1=1+(k-1)×1=k.所以S100=(a1+a3+…+a99)+(a2+a4+a6+a8+…+a100)=×50+25=1275+25=1300.答案: 13007.(2016·全國卷乙)設(shè)等比數(shù)列{an}滿足a1+a3=10,a2+a4=5,則a1a2…an的最大值為________.解析: 設(shè)等比數(shù)列{an}的公比為q,則由a1+a3=10,a2+a4=q(a1+a3)=5,知q=.又a1+a1q2=10,∴a1=8.故a1a2…an=aq1+2+…+(n-1)=23n·=23n-+=2-+n.62018屆高三數(shù)學(xué)二輪復(fù)習(xí)專題集訓(xùn)記t=-+=-(n
6、2-7n),結(jié)合n∈N*可知n=3或4時(shí),t有最大值6.又y=2t為增函數(shù),從而a1a2…an的最大值為26=64.答案: 648.設(shè)某數(shù)列的前n項(xiàng)和為Sn,若為常數(shù),則稱該數(shù)列為“和諧數(shù)列”.若一個(gè)首項(xiàng)為1,公差為d(d≠0)的等差數(shù)列{an}為“和諧數(shù)列”,則該等差數(shù)列的公差d=________.解析: 由=k(k為常數(shù)),且a1=1,得n+n(n-1)d=k,即2+(n-1)d=4k+2k(2n-1)d,整理得,(4k-1)dn+(2k-1)(2-d)=0.∵對(duì)任意正整數(shù)n,上式恒成立,∴得∴數(shù)列{an}的公差為2.答案: 29.已知等比數(shù)列{an}的前
7、n項(xiàng)和為Sn,且6Sn=3n+1+a(n∈N*).(1)求a的值及數(shù)列{an}的通項(xiàng)公式;(2)若bn=(1-an)log3(a·an+1),求數(shù)列的前n項(xiàng)和Tn.解析: (1)∵6Sn=3n+1+a(n∈N*),∴當(dāng)n=1時(shí),6S1=6a1=9+a,當(dāng)n≥2時(shí),6an=6(Sn-Sn-1)=2×3n,即an=3n-1,∵{an}是等比數(shù)列,∴a1=1,則9+a=6,得a=-3,∴數(shù)列{an}的通項(xiàng)公式為an=3n-1(n∈N*).(2)由(1)得bn=(1-an)log3(a·an+1)=(3n-2)(3n+1),∴Tn=++…+=++…+==.10.設(shè)數(shù)列
8、{an}的前n項(xiàng)和為Sn,且點(diǎn)(n,S