《2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課后限時集訓(xùn)31 數(shù)列求和 理(含解析)新人教A版》由會員分享,可在線閱讀,更多相關(guān)《2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課后限時集訓(xùn)31 數(shù)列求和 理(含解析)新人教A版(6頁珍藏版)》請在裝配圖網(wǎng)上搜索。
1、課后限時集訓(xùn)(三十一)數(shù)列求和(建議用時:60分鐘)A組基礎(chǔ)達(dá)標(biāo)一、選擇題1等差數(shù)列an中,已知公差d,且a1a3a9950,則a2a4a100()A50B75C100D125Ban是等差數(shù)列,公差d,a2a4a100(a1a3a99)50d505075.211的值為()A18 B20C22 D18B設(shè)an12.則原式a1a2a11222222220.3已知等比數(shù)列an的前n項和為Sn,若S37,S663,則數(shù)列nan的前n項和為()A3(n1)2n B3(n1)2nC1(n1)2n D1(n1)2nD設(shè)等比數(shù)列an的公比為q,S37,S663,q1,解得an2n1,nann2n1,設(shè)數(shù)列na
2、n的前n項和為Tn,Tn122322423(n1)2n2n2n1,2Tn2222323424(n1)2n1n2n,兩式相減得Tn1222232n1n2n2n1n2n(1n)2n1,Tn1(n1)2n,故選D.4(2019湘潭模擬)已知Sn為數(shù)列an的前n項和,若a12且Sn12Sn,設(shè)bnlog2an,則的值是()A. B.C. D.B由Sn12Sn可知,數(shù)列Sn是首項為S1a12,公比為2的等比數(shù)列,所以Sn2n.當(dāng)n2時,anSnSn12n2n12n1.bnlog2an當(dāng)n2時,所以112.故選B.5已知函數(shù)f(x)xa的圖象過點(diǎn)(4,2),令an,nN*,記數(shù)列an的前n項和為Sn,則S
3、2 019()A.1 B.1C.1 D.1C由f(4)2得4a2,解得a,則f(x)x.an,S2 019a1a2a3a2 019()()()()1.二、填空題6設(shè)數(shù)列an 的前n項和為Sn,且ansin,nN*,則S2 018_.1ansin,nN*,顯然每連續(xù)四項的和為0.S2 018S4504a2 017a2 0180101.7已知數(shù)列an滿足a11,an1an2n(nN*),則S2 018_.321 0093數(shù)列an滿足a11,an1an2n,n1時,a22,n2時,anan12n1,由得2,數(shù)列an的奇數(shù)項、偶數(shù)項分別成等比數(shù)列,S2 018321 0093.8設(shè)Sn1357(1)n
4、1(2n1)(nN*),則Sn_.(1)n1n當(dāng)n為偶數(shù)時,Sn(13)(57)(2n3)(2n1)(2222)2n.當(dāng)n為奇數(shù)時,Sn(13)(57)(2n5)(2n3)(2n1)2(2n1)n.Sn(1)n1n.三、解答題9(2018開封一模)已知數(shù)列an滿足a11,且2nan12(n1)ann(n1)(1)求數(shù)列an的通項公式;(2)若bn,求數(shù)列bn的前n項和Sn.解(1)由已知可得,數(shù)列是以1為首項,為公差的等差數(shù)列,an.(2)bn,bn2,Sn22.10(2018洛陽一模)已知各項均不為零的數(shù)列an的前n項和為Sn,且對任意的nN*,滿足Sna1(an1)(1)求數(shù)列an的通項公
5、式;(2)設(shè)數(shù)列bn滿足anbnlog2an,數(shù)列bn的前n項和為Tn,求證:Tn.解(1)當(dāng)n1時,a1S1a1(a11)aa1,a10,a14.Sn(an1),當(dāng)n2時,Sn1(an11),兩式相減得an4an1(n2),數(shù)列an是首項為4,公比為4的等比數(shù)列,an4n.(2)證明:anbnlog2an2n,bn,Tn,Tn,兩式相減得Tn22.Tn.B組能力提升1(2018石家莊一模)已知函數(shù)f(x)的圖象關(guān)于x1對稱,且f(x)在(1,)上單調(diào),若數(shù)列an是公差不為0的等差數(shù)列,且f(a50)f(a51),則an的前100項的和為()A200B100C0D50B因為函數(shù)f(x)的圖象關(guān)
6、于x1對稱,又函數(shù)f(x)在(1,)上單調(diào),數(shù)列an是公差不為0的等差數(shù)列,且f(a50)f(a51),所以a50a512,所以S10050(a50a51)100,故選B.2(2019鄭州模擬)在數(shù)列an中,若對任意的nN*均有anan1an2為定值,且a72,a93,a984,則數(shù)列an的前100項的和S100()A132 B299 C68 D99B因為在數(shù)列an中,若對任意的nN*均有anan1an2為定值,所以an3an,即數(shù)列an中各項是以3為周期呈周期變化的因為a72,a93,a98a3308a84,所以a1a2a3a7a8a92439,所以S10033(a1a2a3)a100339
7、a7299,故選B.3(2019濟(jì)南模擬)如圖,將平面直角坐標(biāo)系中的格點(diǎn)(橫、縱坐標(biāo)均為整數(shù)的點(diǎn))按如下規(guī)則標(biāo)上標(biāo)簽:原點(diǎn)處標(biāo)數(shù)字0,記為a0;點(diǎn)(1,0)處標(biāo)數(shù)字1,記為a1;點(diǎn)(1,1)處標(biāo)數(shù)字0,記為a2;點(diǎn)(0,1)處標(biāo)數(shù)字1,記為a3;點(diǎn)(1,1)處標(biāo)數(shù)字2,記為a4;點(diǎn)(1,0)處標(biāo)數(shù)字1,記為a5;點(diǎn)(1,1)處標(biāo)數(shù)字0,記為a6;點(diǎn)(0,1)處標(biāo)數(shù)字1,記為a7;以此類推,格點(diǎn)坐標(biāo)為(i,j)的點(diǎn)處所標(biāo)的數(shù)字為ij(i,j均為整數(shù)),記Sna1a2an,則S2 018_.249設(shè)an的坐標(biāo)為(x,y),則anxy.第一圈從點(diǎn)(1,0)到點(diǎn)(1,1)共8個點(diǎn),由對稱性可知a1a2
8、a80;第二圈從點(diǎn)(2,1)到點(diǎn)(2,2)共16個點(diǎn),由對稱性可知a9a10a240,以此類推,可得第n圈的8n個點(diǎn)對應(yīng)的這8n項的和也為0.設(shè)a2 018在第k圈,則8168k4k(k1),由此可知前22圈共有2 024個數(shù),故S2 0240,則S2 018S2 024(a2 024a2 023a2 019),a2 024所在點(diǎn)的坐標(biāo)為(22,22),a2 0242222,a2 023所在點(diǎn)的坐標(biāo)為(21,22),a2 0232122,以此類推,可得a2 0222022,a2 0211922,a2 0201822,a2 0191722,所以a2 024a2 023a2 019249,故S2
9、018249.4各項均為正數(shù)的數(shù)列an的首項a1,前n項和為Sn,且Sn1Sna.(1)求an的通項公式;(2)若數(shù)列bn滿足bnnan,求bn的前n項和Tn.解(1)因為Sn1Sna,所以當(dāng)n2時,SnSn1a,得,an1anaa,即an1an(an1an)(an1an),因為an的各項均為正數(shù),所以an1an0,且0,所以an1an(n2)由知,S2S1a,即2a1a2a,又a1,所以a2.所以a2a1.故an1an(nN*),所以數(shù)列an是首項為,公差為的等差數(shù)列,所以an(n1).(2)由(1)得an,所以bnnn1,所以Tn1232(n1)n2nn1,Tn2233(n1)n1nn,得(1)Tn12n1nn,當(dāng)0且1時,(1)Tnnn,得Tn,當(dāng)1時,由得Tn123(n1)n.綜上,數(shù)列bn的前n項和Tn- 6 -