2018-2019版高中数学 第二章 数列 2.5.3 习题课——数列求和练习 新人教A版必修5.docVIP

习题课数列求和课后篇巩固探究A组1.已知数列an的前n项和为Sn,若an=1n(n+2),则S5等于()A.B.5021C.2521D.2542解析因为an=1n(n+2)=121n-1n+2,所以S5=a1+a2+a3+a4+a5=121-13+12-14+13-15+14-16+15-17=2542.答案D2.已知数列an的通项公式an=1n+n+1,若该数列的前k项之和等于9,则k等于()A.99B.98C.97D.96解析因为an=1n+n+1=n+1-n,所以其前n项和Sn=(2-1)+(3-2)+(n+1-n)=n+1-1.令k+1-1=9,解得k=99.答案A3.数列1,2,3,42716,的前n项和为()A. (n2+n-2)+32nB. n(n+1)+1-32nC. (n2-n+2)-32nD. n(n+1)+31-32n解析数列的前n项和为1+2+3+n+1232n-1=(1+2+3+n)+12+34+98+1232n-1=n(n+1)2+121-32n1-32=n(n+1)2+32n-1= (n2+n-2) +32n,故选A.答案A4.已知an为等比数列,bn为等差数列,且b1=0,cn=an+bn,若数列cn是1,1,2,则数列cn的前10项和为()A.978B.557C.467D.979解析由题意可得a1=1,设数列an的公比为q,数列bn的公差为d,则q+d=1,q2+2d=2,q2-2q=0.q0,q=2,d=-1.an=2n-1,bn=(n-1)(-1)=1-n,cn=2n-1+1-n.设数列cn的前n项和为Sn,则S10=20+0+21-1+29-9=(20+21+29)-(1+2+9)=1-2101-2-9(9+1)2=1 023-45=978.答案A5.已知数列an满足a1=1,a2=2,an+2=1+cos2n2an+sin2n2,则该数列的前18项和为()A.2 101B.2 012C.1 012D.1 067解析由题意可得a3=a1+1,a5=a3+1=a1+2,所以奇数项组成以公差为1,首项为1的等差数列,共有9项,因此S奇=9(1+9)2=45.偶数项a4=2a2,a6=2a4=22a2,因此偶数项组成以2为首项,2为公比的等比数列,共有9项,所以S偶=2(1-29)1-2=-2+210=1 022.故数列an的前18项和为1 022+45=1 067.答案D6.已知数列an的通项公式an=2n-12n,则其前n项和为.解析数列an的前n项和Sn=21-12+22-122+2n-12n=2(1+2+n)-12+122+12n=2n(n+1)2-121-(12)n1-12=n2+n+12n-1.答案n2+n+12n-17.数列112+3,122+6,132+9,142+12,的前n项和等于.解析an=1n2+3n=131n-1n+3,Sn=131-14+12-15+13-16+1n-1-1n+2+1n-1n+3=131+12+13-1n+1-1n+2-1n+3=1118-3n2+12n+113(n+1)(n+2)(n+3).答案1118-3n2+12n+113(n+1)(n+2)(n+3)8.已知等差数列an的前n项和Sn满足S3=0,S5=-5.(1)求an的通项公式;(2)求数列1a2n-1a2n+1的前n项和Tn.解(1)设an的公差为d,则Sn=na1+n(n-1)2d.由已知可得3a1+3d=0,5a1+10d=-5,解得a1=1,d=-1.故an的通项公式为an=2-n.(2)由(1)知1a2n-1a2n+1=1(3-2n)(1-2n)=1(2n-3)(2n-1)=1212n-3-12n-1,从而数列1a2n-1a2n+1的前n项和为Tn=121-1-11+11-13+12n-3-12n-1=n1-2n.9.导学号04994055(2017辽宁统考)已知等差数列an的公差为2,且a1,a1+a2,2(a1+a4)成等比数列.(1)求数列an的通项公式;(2)设数列an2n-1的前n项和为Sn,求证:Sn0,Sn=6-2n+32n-16.B组1.已知数列an的通项公式an=(-1)n-1n2,则其前n项和为()A.(-1)n-1n(n+1)2B.(-1)nn(n+1)2C.n(n+1)2D.-n(n+1)2解析依题意Sn=12-22+32-42+(-1)n-1n2.当n为偶数时,Sn=12-22+32-42+-n2=(12-22)+(32-42)+(n-1)2-n2=- 1+2+3+4+(n-1)+n=-n(n+1)2.当n为奇数时,Sn=12-22+32-42+-(n-1)2+n2=Sn-1+n2=-n(n-1)2+n2=n(n+1)2.Sn=(-1)n-1n(n+1)2.故选A.答案A2.已知数列an为12,13+23,14+24+34,15+25+35+45,则数列bn=1anan+1的前n项和Sn为()A.41-1n+1B.412-1n+1C.1-1n+1D.12-1n+1解析an=1+2+3+nn+1=n(n+1)2n+1=n2,bn=1anan+1=4n(n+1)=41n-1n+1.Sn=41-12+12-13+13-14+1n-1n+1=41-1n+1.答案A3.已知Sn是数列an的前n项和,a1=1,a2=2,a3=3,数列an+an+1+an+2是公差为2的等差数列,则S25=()A.232B.233C.234D.235解析令bn=an+an+1+an+2,则b1=1+2+3=6,由题意知bn=6+2(n-1)=2n+4.因为S25=a1+a2+a3+a25=a1+b2+b5+b23,而b2,b5,b23构成公差为6的等差数列,且b2=8,于是S25=1+88+8726=233.答案B4.数列11,103,1 005,10 007,的前n项和Sn=.解析因为数列的通项公式为an=10n+(2n-1),所以Sn=(10+1)+(102+3)+(10n+2n-1)=(10+102+10n)+1+3+(2n-1)=10(1-10n)1-10+n(1+2n-1)2=109(10n-1)+n2.答案109(10n-1)+n25.数列1,1+2,1+2+22,1+2+22+2n-1,的前n项和等于.解析数列的通项为an=1+2+22+2n-1,因为an=1+2+22+2n-1=1-2n1-2=2n-1,所以该数列的前n项和Sn=(21-1)+(22-1)+(2n-1)=(2+22+2n)-n=2(1-2n)1-2-n=2n+1-n-2.答案2n+1-n-26.已知数列an的前n项和为Sn=3n2-2n,而bn=3anan+1,Tn是数列bn的前n项和,则使得Tnm20对所有nN*都成立的最小正整数m等于.解析由Sn=3n2-2n,得an=6n-5.bn=3anan+1=3(6n-5)(6n+1)=1216n-5-16n+1,Tn=121-17+17-113+16n-5-16n+1x=121-16n+1.121-16n+112,要使121-16n+1m20对nN*成立,需有m2012,即m10,故符合条件的最小正整数为10.答案107.已知递增数列an的前n项和为Sn,且满足Sn=12(an2+n).(1)求a1及数列an的通项公式;(2)设cn=1an+12-1,n为奇数,32an-1+1,n为偶数,求数列cn的前20项和T20.解(1)当n=1时,a1=S1=12(a12+1),解得a1=1.当n2时,Sn-1=12(an-12+n-1),an=Sn-Sn-1=12(an2-an-12+1),解得 an-an-1=1或an+an-1=1(n2).因为an为递增数列,所以an-an-1=1,an是首项为1,公差为1的等差数列,所以an=n.(2)由题意,知cn=1(n+1)2-1,n为奇数,32n-1+1,n为偶数,所以T20=122-1+142-1+1202-1+3(21+23+219)+10=113+135+11921+32(1-410)1-4+10=1211-13+13-15+119-121+2(410-1)+10=1021+221+8=221+17821.8.导学号04994056已知数列an的前n项和为Sn,S1=1,S2=2,当n2时,Sn+1-Sn-1=2n.(1)求证:an+2-an=2n(nN*);(2)求数列an的通项公式;(3)设Tn=a1+a2+122a3+12n-1an,求Tn.(1)证明当n2时,因为an+1+an=2n,an+2+an+1=2n+1,所以an+2-an=2n.又因为a1=1,a2=1,a3=3,所以a3-a1=2,所以an+2-an=2n(nN*).(2)解当n为奇数时,an-a1=(an-an-2)+(an-2-an-4)+(a5-a3)+(a3-a1)=2n-2+2n-4+23+2=21-4n-121-4=23(2n-1-1),所以an=2n3+13.同理,当n为偶数时,an=2n3-1