1、课时规范练31数列求和课时规范练A册第20页基础巩固组1.(2019广东广州调研)数列112,314,518,7116,(2n-1)+12n,的前n项和Sn的值等于() A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案A解析该数列的通项公式为an=(2n-1)+12n,则Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.(2019广东深圳调研)已知函数f(n)=n2,n为奇数,-n2,n为偶数,且an=f(n)+f(n+1),则a1+a2+a3+a100等于()A.0B.100C.-100D.10 200答案B解析由题
2、意,得a1+a2+a3+a100=12-22-22+32+32-42-42+52+992-1002-1002+1012=-(1+2)+(3+2)-(4+3)+-(99+100)+(101+100)=-(1+2+99+100)+(2+3+100+101)=-50101+50103=100.故选B.3.(2019河南开封调研)已知数列an满足a1=1,an+1an=2n(nN+),则S2 018等于()A.22 018-1B.321 009-3C.321 009-1D.321 008-2答案B解析a1=1,a2=2a1=2,又an+2an+1an+1an=2,an+2an=2.a1,a3,a5,成
3、等比数列;a2,a4,a6,成等比数列,S2 018=a1+a2+a3+a4+a5+a6+a2 017+a2 018=(a1+a3+a5+a2 017)+(a2+a4+a6+a2 018)=1-21 0091-2+2(1-21 009)1-2=321 009-3.故选B.4.(2017全国2,理15)等差数列an的前n项和为Sn,a3=3,S4=10,则k=1n1Sk=.答案2nn+1解析设等差数列的首项为a1,公差为d,由题意可知a1+2d=3,4a1+432d=10,解得a1=1,d=1.所以Sn=na1+n(n-1)2d=n(1+n)2.所以1Sn=2n(n+1)=21n-1n+1.所以
4、k=1n1Sk=21-12+12-13+1n-1n+1=21-1n+1=2nn+1.5.(2019安徽合肥一模,15)设等差数列an满足a2=5,a6+a8=30,则数列1an2-1的前n项的和等于.答案n4(n+1)解析设等差数列an的公差为d,由等差数列的性质可得a6+a8=30=2a7,a7=15,a7-a2=5d,即15=5+5d,d=2,an=a2+(n-2)d=2n+1,1an2-1=14n(n+1)=141n-1n+1,前n项和为141-12+12-13+1n-1n+1=141-1n+1=n4(n+1).6.(2019山东淄博一模,17)已知在等比数列an中,a1=2,且a1,a
5、2,a3-2成等差数列.(1)求数列an的通项公式;(2)若数列bn满足:bn=1an+2log2an-1,求数列bn的前n项和Sn.解(1)设等比数列an的公比为q,a1,a2,a3-2成等差数列,2a2=a1+(a3-2)=2+(a3-2)=a3,q=a3a2=2,an=a1qn-1=2n(nN+).(2)由(1)及bn=1an+2log2an-1,可知bn=12n+2log22n-1=12n+2n-1,Sn=12+1+122+3+123+5+12n+(2n-1)=12+122+123+12n+1+3+5+(2n-1)=121-12n1-12+n1+(2n-1)2=n2-12n+1(nN+
6、).7.(2019山东实验等四校联考,17)已知数列an的前n项和Sn满足Sn=Sn-1+1(n2,nN),且a1=1.(1)求数列an的通项公式;(2)记bn=1anan+1,Tn为数列bn的前n项和,求使Tn2n成立的n的最小值.解(1)由已知Sn=Sn-1+1,得Sn-Sn-1=1,所以数列Sn为等差数列,且S1=1.Sn=n,即Sn=n2,当n2时,an=Sn-Sn-1=n2-(n-1)2=2n-1,又a1=1也满足上式,an=2n-1.(2)由(1)知,bn=1(2n-1)(2n+1)=1212n-1-12n+1,Tn=121-13+13-15+12n-1-12n+1=121-12n
7、+1=n2n+1,由Tn2n有n24n+2,有(n-2)26,所以n5,n的最小值为5.综合提升组8.(2019广东珠海一中等六校联考)已知数列an满足a1=1,且对于任意的nN+都有an+1=an+a1+n,则1a1+1a2+1a2 017等于()A.2 0162 017B.4 0322 017C.2 0172 018D.4 0342 018答案D解析由题意可得an+1-an=n+1,则a1=1,a2-a1=2,a3-a2=3,an-an-1=n,以上各式相加可得an=n(n+1)2,则1an=21n-1n+1,1a1+1a2+1a2 017=21-12+12-13+12 017-12 01
8、8=4 0342 018.9.(2019湖南师大附中模拟,12)设数列an的前n项和为Sn,且a1=1,an=Snn+2(n-1),则数列1Sn+3n的前10项的和是()A.290B.920C.511D.1011答案C解析由an=Snn+2(n-1),得Sn=nan-2n(n-1),当n2时,an=Sn-Sn-1=nan-(n-1)an-1-4(n-1),整理得an-an-1=4,所以数列an是公差为4的等差数列,又a1=1,所以an=4n-3,则Sn+3n=n(a1+an)2+3n=2n(n+1),1Sn+3n=12n(n+1)=121n-1n+1,数列1Sn+3n的前10项的和S=121-
9、12+12-13+110-111=121-111=511,故选C.10.(2019衡水联考)已知数列an与bn的前n项和分别为Sn,Tn,且an0,6Sn=an2+3an,nN+,bn=2an(2an-1)(2an+1-1),若任意nN+,kTn恒成立,则k的最小值是.答案149解析当n=1时,6a1=a12+3a1,解得a1=3或a1=0.由an0,得a1=3.由6Sn=an2+3an,得6Sn+1=an+12+3an+1.两式相减得6an+1=an+12-an2+3an+1-3an.所以(an+1+an)(an+1-an-3)=0.因为an0,所以an+1+an0,an+1-an=3.即数
10、列an是以3为首项,3为公差的等差数列,所以an=3+3(n-1)=3n.所以bn=2an(2an-1)(2an+1-1)=8n(8n-1)(8n+1-1)=1718n-1-18n+1-1.所以Tn=1718-1-182-1+182-1-183-1+18n-1-18n+1-1=1717-18n+1-1Tn恒成立,只需k149.11.(2019山东淄博实验中学期末,17)已知等差数列an的公差d0,其前n项和为Sn,且S5=20,a3,a5,a8成等比数列.(1)求数列an的通项公式;(2)令bn=1anan+1+n,求数列bn的前n项和Tn.解(1)因为S5=5(a1+a5)2=20,即a1+
11、a5=8,a3=4,即a1+2d=4.因为a3,a5,a8为等比数列,即a52=a3a8.所以(a1+4d)2=(a1+2d)(a1+7d),化简得a1=2d.联立和得a1=2,d=1,所以an=n+1.(2)由(1)及bn=1anan+1+n,可知bn=1anan+1+n=1(n+1)(n+2)+n=1n+1-1n+2+n,所以Tn=12-13+1+13-14+2+14-15+3+1n+1-1n+2+n=12-13+13-14+14-15+1n+1-1n+2+(1+2+3+n)=12-1n+2+n(n+1)2=n2(n+2)+n(n+1)2.12.(2019贵州贵阳一模)已知数列an的前n项
12、和是Sn,且Sn+12an=1(nN+).(1)求数列an的通项公式;(2)设bn=log13(1-Sn+1)(nN+),令Tn=1b1b2+1b2b3+1bnbn+1,求Tn.解(1)当n=1时,a1=S1,由S1+12a1=1,得a1=23.当n2时,Sn=1-12an,Sn-1=1-12an-1,则Sn-Sn-1=12(an-1-an),即an=12(an-1-an),所以an=13an-1(n2).故数列an是以23为首项,13为公比的等比数列.故an=2313n-1=213n(nN+).(2)因为1-Sn=12an=13n.所以bn=log13(1-Sn+1)=log1313n+1=
13、n+1.因为1bnbn+1=1(n+1)(n+2)=1n+1-1n+2,所以Tn=1b1b2+1b2b3+1bnbn+1=12-13+13-14+1n+1-1n+2=12-1n+2=n2(n+2).创新应用组13.(2019河南重点学校月考)已知数列an中,a1=1,an-1-an=2anan-1(n2).(1)求数列an的通项公式;(2)设bn=an2n+1,数列bn的前n项和为Sn,证明:对任意的nN+,都有13Sn0,2+1x2,所以f(x)=x2x+1=12+1x12.所以13f(x)12.即对任意的nN+,都有13Sn0,前n项和为Sn,若an=Sn+Sn-1(nN+,且n2).(1
14、)求数列an的通项公式;(2)记cn=an2an,求数列cn的前n项和Tn.解(1)在数列an中,an=Sn-Sn-1,又有an=Sn+Sn-1(nN+,且n2),所以an=Sn-Sn-1=(Sn+Sn-1)(Sn-Sn-1)=an(Sn-Sn-1),所以Sn-Sn-1=1,所以数列Sn是以S1=a1=1为首项,公差为1的等差数列,所以Sn=1+(n-1)=n,即Sn=n2.当n=1时,a1=S1=1,当n2时,an=Sn-Sn-1=n2-(n-1)2=2n-1,a1=21-1=1也满足上式,所以an的通项公式为an=2n-1.(2)由(1)知cn=an2an=(2n-1)22n-1,Tn=2
15、1+323+525+(2n-3)22n-3+(2n-1)22n-1,4Tn=23+325+527+(2n-3)22n-1+(2n-1)22n+1.-得-3Tn=21+223+225+227+222n-1-(2n-1)22n+1=5-6n322n+1-103,即Tn=6n-5922n+1+109.15.(2019四川百校模拟冲刺改编)定义在0,+)上的函数f(x)满足:当0x2时,f(x)=2x-x2;当x2时,f(x)=3f(x-2).记函数f(x)的极大值点从小到大依次记为a1,a2,an,并记相应的极大值为b1,b2,bn,.(1)求数列an,bn的通项公式;(2)设S=a1b1+a2b2+a20b20,求S的值(不必求出具体的数值).解(1)由题意当0x2时,f(x)=2x-x2=-(x-1)2+1,极大值点为1,极大值为1,当x2时,f(x)=3f(x-2).则极大值点形成首项为1,公差为2的等差数列,极大值形成首项为1,公比为3的等比数列,故an=2n-1,bn=3n-1,故anbn=(2n-1)3n-1.(2)由S=a1b1+a2b2+a20b20=11+331+532+39319,则3S=131+332+39320,两式相减得-2S=1+2(31+32+319)-39320=1+23(1-319)1-3-39320=-2-38320,S=19320+1.
