1、2.5等比数列的前n项和第1课时等比数列的前n项和课后篇巩固提升基础巩固1.已知数列an的通项公式是an=2n,Sn是数列an的前n项和,则S10等于()A.10B.210C.a10-2D.211-2解析an+1an=2n+12n=2,数列an是公比为2的等比数列,且a1=2.S10=2(1-210)1-2=211-2.答案D2.在等比数列an中,a2=9,a5=243,则an的前4项和为()A.81B.120C.168D.192解析因为a5a2=27=q3,所以q=3,a1=a2q=3,S4=3(1-34)1-3=120.答案B3.已知等比数列an的前n项和为Sn,且a1+a3=52,a2+
2、a4=54,则Snan=()A.4n-1B.4n-1C.2n-1D.2n-1解析设公比为q,则q=a2+a4a1+a3=12,于是a1+14a1=52,因此a1=2,于是Sn=21-12n1-12=41-12n,而an=212n-1=12n-2,于是Snan=41-12n12n-2=2n-1.答案D4.在14与78之间插入n个数组成一个等比数列,若各项总和为778,则此数列的项数为()A.4B.5C.6D.7解析设a1=14,an+2=78,则Sn+2=14-78q1-q=778,解得q=-12.所以an+2=14-12n+1=78,解得n=3.故该数列共5项.答案B5.各项均为正数的等比数列
3、an的前n项和为Sn,若S10=20,S30=140,则S40=()A.280B.300C.320D.340解析设等比数列an的公比为q1.S10=10,S30=70,a11-q(1-q10)=20,a11-q(1-q30)=140,联立解得q10=2,a11-q=-20,则S40=a11-q(1-q40)=-20(1-24)=300,故选B.答案B6.对于等比数列an,若a1=5,q=2,Sn=35,则an=.解析由Sn=a1-anq1-q,得an=a1-(1-q)Snq=5+352=20.答案207.在等比数列an中,设前n项和为Sn,若a3=2S2+1,a4=2S3+1,则公比q=.解析
4、因为a3=2S2+1,a4=2S3+1,两式相减,得a4-a3=2a3,即a4=3a3,所以q=a4a3=3.答案38.已知等比数列an是递减数列,Sn是an的前n项和,若a1,a2是方程2x2-3x+1=0的两个根,则S5=.解析a1,a2是方程2x2-3x+1=0的两根,且等比数列an是递减数列,a1=1,a2=12,则公比q=12,S5=a1(1-q5)1-q=1-1251-12=3116.答案31169.已知等比数列an满足a3=12,a8=38,记其前n项和为Sn.(1)求数列an的通项公式an;(2)若Sn=93,求n.解(1)设等比数列an的公比为q,则a3=a1q2=12,a8
5、=a1q7=38,解得a1=48,q=12,所以an=a1qn-1=4812n-1.(2)Sn=a1(1-qn)1-q=481-12n1-12=961-12n.由Sn=93,得961-12n=93,解得n=5.10.已知等差数列an的首项为a,公差为b,方程ax2-3x+2=0的解为1和b(b1).(1)求数列an的通项公式;(2)若数列an满足bn=an2n,求数列bn的前n项和Tn.解(1)因为方程ax2-3x+2=0的两根为x1=1,x2=b,可得a-3+2=0,ab2-3b+2=0,解得a=1,b=2.所以an=2n-1.(2)由(1)得bn=(2n-1)2n,所以Tn=b1+b2+b
6、n=12+322+(2n-1)2n,2Tn=122+323+(2n-3)2n+(2n-1)2n+1,由-,得-Tn=12+222+223+22n-(2n-1)2n+1=2(2+22+23+2n)-(2n-1)2n+1-2=22(1-2n)1-2-(2n-1)2n+1-2=(3-2n)2n+1-6.所以Tn=(2n-3)2n+1+6.能力提升1.等比数列an的前n项和为Sn,若S2n=3(a1+a3+a2n-1),a1a2a3=8,则Sn=()A.2n-1B.2n-1-1C.2n+1-1D.2n+1解析显然q1,由已知,得a1(1-q2n)1-q=3a1(1-q2n)1-q2,整理,得q=2.因
7、为a1a2a3=8,所以a23=8,所以a2=2,从而a1=1.于是Sn=1-2n1-2=2n-1.答案A2.已知数列an是首项为1的等比数列,Sn是an的前n项和,且9S3=S6,则数列1an的前5项和为()A.158或5B.3116或5C.3116D.158解析由题意易知公比q1.由9S3=S6,得9a1(1-q3)1-q=a1(1-q6)1-q,解得q=2.所以1an是首项为1,公比为12的等比数列.所以其前5项和为S5=11-1251-12=3116.答案C3.在等比数列an中,a1+a2+a5=27,1a1+1a2+1a5=3,则a3=()A.9B.9C.3D.3解析设公比为q,则由
8、已知可得a1(1-q5)1-q=27,1a11-1q51-1q=3,两式相除,得a12q4=9,即a32=9,所以a3=3.答案C4.若等比数列an的前n项和为Sn,且S1,S3,S2成等差数列,则an的公比q=.解析由题意,得a1+(a1+a1q)=2(a1+a1q+a1q2),又a10,q0,故q=-12.答案-125.1+322+423+n2n-1+n+12n=.解析设Sn=1+322+423+n2n-1+n+12n,则12Sn=222+323+424+n2n+n+12n+1,两式相减,得12Sn=1+122+123+12n-n+12n+1=12+121-12n1-12-n+12n+1=
9、32-12n-n+12n+1.所以Sn=3-n+32n.答案3-n+32n6.若等比数列an的前n项和为Sn,且S3+S6=2S9,则公比q等于.解析若q=1,S3+S6=3a1+6a1=9a12S9.q1,a1(1-q3)1-q+a1(1-q6)1-q=2a1(1-q9)1-q,即 2q9-q6-q3=0,q3(2q6-q3-1)=0.q0,2q6-q3-1=0,(q3-1)(2q3+1)=0,q3=-12或q3=1(舍),q=-342.答案-3427.已知等比数列an的各项均为正数,且2a1+3a2=1,a52=9a4a8.(1)求数列an的通项公式;(2)设bn=an-an-1,求数列b
10、n的前n项和Sn.解(1)设an的公比为q,则由a52=9a4a8,得(a1q4)2=9a1q3a1q7,即a12q8=9a12q10,因此q2=19.因为an的各项均为正数,所以q0,所以q=13.又因为2a1+3a2=1,所以2a1+3a113=1,解得a1=13,故an=1313n-1,即an=13n.(2)由(1)得bn=an-an-1=13n-13n-1=-2313n-1,所以bn是首项为-23,公比为13的等比数列,因此其前n项和Sn=-231-13n1-13=13n-1.8.已知数列an的前n项和Sn=an+n2-1,数列bn满足3nbn+1=(n+1)an+1-nan,且b1=
11、3.(1)求an,bn;(2)设Tn为数列bn的前n项和,求Tn.解(1)当n2时,Sn=an+n2-1,Sn-1=an-1+(n-1)2-1,两式相减,得an=an-an-1+2n-1,an-1=2n-1.an=2n+1.3nbn+1=(n+1)(2n+3)-n(2n+1)=4n+3.bn+1=4n+33n,当n2时,bn=4n-13n-1.又b1=3适合上式,bn=4n-13n-1.(2)由(1)知bn=4n-13n-1,Tn=31+73+1132+4n-53n-2+4n-13n-1,13Tn=33+732+1133+4n-53n-1+4n-13n,-,得23Tn=3+43+432+43n-1-4n-13n=3+4131-13n-11-13-4n-13n=5-5+4n3n.Tn=152-4n+523n-1.