（新课改地区）2021版高考数学一轮复习 核心素养测评三十四 数列求和 新人教B版

核心素养测评三十四 数列求和(30分钟60分)一、选择题(每小题5分,共25分)1.数列an的通项公式是an=1n+n+1,若前n项和为10,则项数n为()a.120b.99c.11d.121【解析】选a.an=1n+n+1=n+1-n(n+1+n)(n+1-n)=n+1-n,所以a1+a2+an=(2-1)+(3-2)+(n+1-n)=n+1-1=10.即n+1=11,所以n+1=121,n=120.2.已知数列an的通项公式是an=2n-315n,则其前20项和为()a.380-351-1519b.400-251-1520c.420-341-1520d.440-451-1520【解析】选c.令数列an的前n项和为sn,则s20=a1+a2+a20=2(1+2+20)-315+152+1520=220(20+1)2-3151-15201-15=420-341-1520.3.在数列an中,an=2n-12n,若an的前n项和sn=32164,则n=()a.3b.4c.5 d.6【解析】选d.由an=2n-12n=1-12n得:sn=n-12+122+12n=n-1-12n,则sn=32164=n-1-12n,将各选项中的值代入验证得n=6.4.(多选)已知数列an:12,13+23,14+24+34,110+210+910,若bn=1anan+1,设数列bn的前n项和为sn,则()a.an=n2b.an=nc.sn=4nn+1d.sn=5nn+1【解析】选ac.由题意得an=1n+1+2n+1+nn+1=1+2+3+nn+1=n2,所以bn=1n2n+12=4n(n+1)=41n-1n+1,所以数列bn的前n项和sn=b1+b2+b3+bn=41-12+12-13+13-14+1n-1n+1=41-1n+1=4nn+1.5.已知数列an满足a1=1,an+1an=2n(nn*),sn是数列an的前n项和,则s2 020=()a.22 020-1b.321 010-3c.321 010-1d.322 020-2【解析】选b.依题意得anan+1=2n,an+1an+2=2n+1,于是有an+1an+2anan+1=2,即an+2an=2,数列a1,a3,a5,a2n-1,是以a1=1为首项,2为公比的等比数列;数列a2,a4,a6,a2n,是以a2=2为首项,2为公比的等比数列,于是有s2 020=(a1+a3+a5+a2 019)+(a2+a4+a6+a2 020)=1-21 0101-2+2(1-21 010)1-2=321 010-3.二、填空题(每小题5分,共15分)6.在数列an中,若an+1+(-1)nan=2n-1,则数列an的前12项和等于_.【解析】由已知an+1+(-1)nan=2n-1,得an+2+(-1)n+1an+1=2n+1,得an+2+an=(-1)n(2n-1)+(2n+1),取n=1,5,9及n=2,6,10,结果相加可得s12=a1+a2+a3+a4+a11+a12=78.答案:787.已知数列an,bn,若b1=0,an=1n(n+1),当n2时,有bn=bn-1+an-1,则b10=_.【解析】由bn=bn-1+an-1得bn-bn-1=an-1,所以b2-b1=a1,b3-b2=a2,bn-bn-1=an-1,所以b2-b1+b3-b2+bn-bn-1=a1+a2+an-1=112+123+1(n-1)n,即bn-b1=a1+a2+an-1=112+123+1(n-1)n=11-12+12-13+1n-1-1n=1-1n=n-1n,又因为b1=0,所以bn=n-1n,所以b10=910.答案:9108.设数列an的通项公式为an=22n-1,令bn=nan,则数列bn的前n项和sn为_.【解析】由bn=nan=n22n-1知,sn=12+223+325+n22n-1,从而22sn=123+225+327+n22n+1,-得(1-22)sn=2+23+25+22n-1-n22n+1,即sn=19(3n-1)22n+1+2.答案: 19(3n-1)22n+1+2三、解答题(每小题10分,共20分)9.(2020兰州模拟)已知数列an的前n项和sn满足2sn=an-1an+2,且an0nn*.(1)求数列an的通项公式.(2)若bn=3n(2n-1)nannn*,记数列bn的前n项和为tn,证明:tn32.【解析】(1)当n=1时,2s1=a1-1a1+2=2a1,因为a10,所以a1=2,当n2时,2an=2sn-sn-1=an-1an+2-an-1-1an-1+2,所以an+an-1an-an-1-1=0,因为an0,所以an-an-1-1=0,所以an-an-1=1,所以an是以a1=2为首项,d=1为公差的等差数列,所以an=n+1nn*.(2)由(1)得an=n+1,所以bn=3n2n-1nn+1=3n+1n+1-3nn,所以tn=b1+b2+bn-1+bn=322-3+333-322+3nn-3n-1n-1+3n+1n+1-3nn=3n+1n+1-3,因为tn+1-tn=3n+2n+2-3n+1n+1=3n+12n+1n+1n+20,所以tn是递增数列,所以tnt1=92-3=32.10.已知数列an的各项均为正数,且an2-2nan-(2n+1)=0,nn*.(1)求数列an的通项公式.(2)若bn=2nan,求数列bn的前n项和tn.【解析】(1)由an2-2nan-(2n+1)=0得an-(2n+1)(an+1)=0,所以an=2n+1或an=-1,又数列an的各项均为正数,负值应舍去,所以an=2n+1,nn*.(2)因为bn=2nan=2n(2n+1),所以tn=23+225+237+2n(2n+1),2tn=223+235+2n(2n-1)+2n+1(2n+1),由-得-tn=6+2(22+23+2n)-2n+1(2n+1)=6+222(1-2n-1)1-2-2n+1(2n+1)=-2+2n+1(1-2n).所以tn=(2n-1)2n+1+2.(15分钟35分)1.(5分)若数列an的通项公式是an=(-1)n(3n-2),则a1+a2+a3+a10=()a.15b.12c.-12d.-15【解析】选a.因为an=(-1)n(3n-2),所以a1+a2+a10=-1+4-7+10-13+16-19+22-25+28=(-1+4)+(-7+10)+(-13+16)+(-19+22)+(-25+28)=35=15.【变式备选】已知数列an的前n项和为sn,通项公式an=n(-1)n+1,则s17=()a.10b.9c.8d.7【解析】选b.s17=1-2+3-4+5-6+15-16+17=1+(-2+3)+(-4+5)+(-14+15)+(-16+17)=1+1+1+1=9.【一题多解】解决本题还可以采用以下方法:选b.s17=1-2+3-4+5-6+15-16+17=(1+3+17)-(2+4+16)=81-72=9.2.(5分)已知等比数列an的首项为32,公比为-12,其前n项和为sn,则sn的最大值为()a.34b.23c.43d.32【解析】选d.因为等比数列an的首项为32,公比为-12,所以sn=321-12n1-12=1-12n,当n取偶数时,sn=1-12n0,解得q=2,所以bn=2n.设等差数列an的公差为d,由b3=a4-2a1可得3d-a1=8,由s11=11b4可得a1+5d=16,联立解得a1=1,d=3,由此可得an=3n-2.所以an的通项公式为an=3n-2,bn的通项公式为bn=2n.(2)设数列a2nbn的前n项和为tn,由a2n=6n-2得tn=42+1022+1623+(6n-2)2n,2tn=422+1023+1624+(6n-8)2n+(6n-2)2n+1,上述两式相减得:-tn=42+622+623+62n-(6n-2)2n+1=12(1-2n)1-2-4-(6n-2)2n+1=-(3n-4)2n+2-16.所以tn=(3n-4)2n+2+16.所以数列a2nbn的前n项和为(3n-4)2n+2+16.1.已知数列an的前n项积为tn,若对n2,nn*,都有tn+1tn-1=2t n2成立,且a1=1,a2=2,则数列an的前10项和为_.【解析】因为tn+1tn-1=2t n2,故tn+1tntntn-1=2,即an+1an=2(n2),而a2a1=2,所以an是首项为1,公比为2的等比数列,故an=2n-1,所以s10=11-2101-2=1 023.答案:1 0232.已知正项数列an中,a1=1,a2=2,2an2=an-12+an+12(n2),bn=1an+an+1,数列bn的前n项和为sn,则s33的值是_.【解析】