2023年高考数学一轮复习 课时规范练31 数列求和（含解析）北师大版 文.docx

1、课时规范练31数列求和1.已知在等比数列an中,a1=2,且a1,a2,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.(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)=121-1

2、2n1-12+n1+(2n-1)2=n2-12n+1.2.已知数列an的前n项和为Sn,且满足2Sn+an-n=0.(1)求证:数列an-12为等比数列;(2)求数列an-n的前n项和Tn.(1)证明:当n=1时,2S1+a1-1=0,解得a1=13.因为2Sn+an-n=0(nN*),当n2时,2Sn-1+an-1-(n-1)=0,-,得3an=an-1+1,即an=13an-1+13,当n2时,an-12an-1-12=13an-1+13-12an-1-12=13,又a1-12=-16,所以an-12是以-16为首项,以13为公比的等比数列.(2)解:由(1)可得an=-1213n+12,

3、所以an-n=-1213n-n+12,所以数列an-n的前n项和Tn=-12131-13n1-13-(n+1)n2+n2,化简得Tn=1413n-1-n22.3.(2021四川成都石室中学高三月考)在数列an中,a1=3,且对任意nN+,都有an+an+22=an+1.(1)设bn=an+1-an,判断数列bn是否为等差数列或等比数列;(2)若a2=5,cn=an,n为奇数,2an-1,n为偶数,求数列cn的前2n项的和S2n.解:(1)由an+an+22=an+1,得an+an+2=2an+1,an+2-an+1=an+1-an,所以数列an是等差数列.当an的公差为零时,bn+1=bn=0

4、,数列bn是等差数列,不是等比数列;当an的公差不为零时,bn+1=bn0,数列bn既是等差数列也是等比数列.(2)若a2=5,由(1)知an+1-an=a2-a1=2,所以数列an是等差数列,且首项为3,公差为2,所以an=3+2(n-1)=2n+1.则cn=2n+1,n为奇数,4n,n为偶数.S2n=S奇+S偶=3+7+11+(4n-1)+(42+44+42n)=(3+4n-1)n2+16(1-16n)1-16=2n2+n+16(16n-1)15.4.(2021云南昆明“三诊一模”检测)已知等差数列an的前n项和为Sn,a3n=3an-2,且S5-S3=4a2.(1)求数列an的通项公式;

5、(2)设数列1Sn的前n项和为Tn,证明:Tn0,所以Tn0,所以解得a2=2,故数列an的公差为a2-a1=1,所以an=1+n-1=n.(2)由(1)可得bn=(-1)n4n(2an-1)(2an+1)=(-1)n4n(2n-1)(2n+1)=(-1)n12n-1+12n+1,所以T2n=-1+13+13+15-15+17+14n-1+14n+1=-1+14n+1=-4n4n+1.7.(2020新高考,18)已知公比大于1的等比数列an满足a2+a4=20,a3=8.(1)求an的通项公式;(2)记bm为an在区间(0,m(mN+)中的项的个数,求数列bm的前100项和S100.解:(1)

6、设an的公比为q.由题设得a1q+a1q3=20,a1q2=8.解得q=12(舍去),q=2.因为a1q2=8,所以a1=2.所以an的通项公式为an=2n.(2)由题设及(1)知b1=0,且当2nm2n+1时,bm=n.所以S100=b1+(b2+b3)+(b4+b5+b6+b7)+(b32+b33+b63)+(b64+b65+b100)=0+12+222+323+424+525+6(100-63)=480.8.(2021河北秦皇岛模拟)已知数列an满足2an+1=an+1,a1=54,bn=an-1.(1)求证:数列bn是等比数列;(2)求数列的前n项和Tn.从条件n+1bn,n+bn,4

7、log2bnlog2bn+1中任选一个,补充到上面的问题中,并给出解答.(1)证明:因为2an+1=an+1,所以2an+1-2=an-1.因为bn=an-1,所以2bn+1=bn,bn+1=12bn.因为b1=a1-1=14,所以数列bn是以14为首项,12为公比的等比数列,bn=12n+1.(2)解:选:因为bn=12n+1,所以n+1bn=(n+1)2n+1,则Tn=222+323+(n+1)2n+1,2Tn=223+324+(n+1)2n+2,Tn=2Tn-Tn=-222-23-24-2n+1+(n+1)2n+2=(n+1)2n+2-23(1-2n-1)1-2-222=(n+1)2n+2+8-2n+2-8=n2n+2,故Tn=n2n+2.选:因为bn=12n+1,所以n+bn=n+12n+1,则Tn=1+14+2+18+3+116+n+12n+1=(1+2+3+n)+14+18+116+12n+1=12n(n+1)+14(1-12n)1-12=n22+n2+12-12n+1,故Tn=n22+n2+12-12n+1.选:因为bn=12n+1,所以4log2bnlog2bn+1=41n+1-1n+2,则Tn=412-13+13-14+1n-1n+1+1n+1-1n+2=412-1n+2=2nn+2,故Tn=2nn+2.