2023年高考数学一轮复习 课时规范练32 数列求和（含解析）新人教A版 理.docx

1、课时规范练32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.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.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

2、的和S2n.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;(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.解:(1)设an的公比为q.由题设得a1q+a1q3

3、=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.(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.(

4、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.