2021届高考数学二轮复习 题型练4 大题专项（二）数列的通项、求和问题 理（含解析）.docx

1、题型练4大题专项(二)数列的通项、求和问题题型练第64页一、解答题1.已知数列an满足a2-a1=1,其前n项和为Sn,当n2时,Sn-1-1,Sn,Sn+1成等差数列.(1)求证an为等差数列;(2)若Sn=0,Sn+1=4,求n.答案:(1)证明当n2时,由Sn-1-1,Sn,Sn+1成等差数列,可知2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n2),则an+1-an=1(n2),又a2-a1=1,故an是公差为1的等差数列.(2)解由(1)知等差数列an的公差为1.由Sn=0,Sn+1=4,得an+1=4,即a1+n=4.由Sn=0,得

2、na1+n(n-1)2=0,即a1+n-12=0,解得n=7.2.(2020全国,理17)设数列an满足a1=3,an+1=3an-4n.(1)计算a2,a3,猜想an的通项公式并加以证明;(2)求数列2nan的前n项和Sn.解:(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1-(2n+3)=3an-(2n+1),an-(2n+1)=3an-1-(2n-1),a2-5=3(a1-3).因为a1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以Sn=32+522+723+(2n+1)2n.从而2Sn=322+523+724+(2n+1)2n+1.-得-Sn

3、=32+222+223+22n-(2n+1)2n+1.所以Sn=(2n-1)2n+1+2.3.已知等比数列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列bn满足b1=1,数列(bn+1-bn)an的前n项和为2n2+n.(1)求q的值;(2)求数列bn的通项公式.解:(1)由a4+2是a3,a5的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列cn的前n项和为Sn,由cn=S1,n=1,Sn-Sn

4、-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1.故bn-bn-1=(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.设Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12n-2+(4n-5)12n-1,所以12Tn=3+412+4122+412n-2-(4n-5)12n-1,因此Tn=14-(4n+3)12n-2,n2,又b1=1,所以bn=15-(4n+3)1

5、2n-2.4.已知等差数列an的前n项和为Sn,公比为q的等比数列bn的首项是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列an,bn的通项公式an,bn;(2)求数列1anan+1+1bnbn+1的前n项和Tn.解:(1)设an公差为d,由题意得a1+2d=8,a1+2q=3,a1+d+2q=6,解得a1=2,d=3,q=12,故an=3n-1,bn=12n.(2)1anan+1+1bnbn+1=131an-1an+1+1bnbn+1=131an-1an+1+22n+1,Tn=1312-15+15-18+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+

6、13(22n+3-8)=1322n+3-13n+2-52.5.已知数列an满足a1=12,且an+1=an-an2(nN*).(1)证明:1anan+12(nN*);(2)设数列an2的前n项和为Sn,证明:12(n+2)Snn12(n+1)(nN*).答案:证明(1)由题意得an+1-an=-an20,即an+1an,故an12.由an=(1-an-1)an-1,得an=(1-an-1)(1-an-2)(1-a1)a10.由0an12,得anan+1=anan-an2=11-an1,2,即1anan+12.(2)由题意得an2=an-an+1,所以Sn=a1-an+1.由1an+1-1an=

7、anan+1和1anan+12,得11an+1-1an2,所以n1an+1-1a12n,因此12(n+1)an+11n+2(nN*).由得12(n+2)Snn12(n+1)(nN*).6.已知等比数列an的前n项和为Sn,且an+1=1+Sn,且a2=2a1.(1)求数列an的通项公式;(2)若bn=anlog2an+(-1)nn,求数列bn的前n项和Hn.解:(1)an+1=1+Sn,当n2时,an=1+Sn-1,an+1=2an(n2).又a2=1+S1=1+a1,a2=2a1,解得a1=1.an=2n-1.(2)由题意可知bn=anlog2an+(-1)nn=(n-1)2n-1+(-1)nn.设数列(n-1)2n-1的前n项和为Tn,则有Tn=020+121+222+(n-1)2n-1,2Tn=021+122+223+(n-1)2n,由-,得Tn=(n-2)2n+2.当n为偶数时,Hn=(n-2)2n+2-1+2-3+-(n-1)+n=(n-2)2n+2+n2=(n-2)2n+n+42.当n为奇数时,Hn=(n-2)2n+2-1+2-3+-(n-1)-n=(n-2)2n+2+n-12-n=(n-2)2n-n-32.故Hn=(n-2)2n+n+42(n为偶数),(n-2)2n-n-32(n为奇数).