1、专题突破练12求数列的通项及前n项和1.(2021湖南长郡中学月考)已知等比数列an的各项均为正数,且2a1+3a2=1,a32=9a2a6.(1)求数列an的通项公式;(2)设bn=log3a1+log3a2+log3an,求数列1bn的前n项和Tn.2.(2021山东威海期末)已知等差数列an的前n项和为Sn,且满足a3=8,S5=2a7.(1)求数列an的通项公式;(2)若数列bn满足bn=ancos n+2n+1,求数列bn的前2n项和T2n.3.(2021东北三省四市联考)已知等差数列an的前n项和为Sn,S5=25,且a3-1,a4+1,a7+3成等比数列.(1)求数列an的通项公
2、式;(2)若bn=(-1)nan+1,Tn是数列bn的前n项和,求T2n.4.(2021陕西西安铁一中月考)已知数列an是公差不为0的等差数列,且a2=3,a1,a2,a5成等比数列.(1)求数列an的通项公式;(2)设Sn为数列an+2的前n项和,bn=1Sn,求数列bn的前n项和Tn.5.(2021广东揭阳检测)已知等差数列an与正项等比数列bn满足a1=b1=3,且b3-a3,20,a5+b2既是等差数列,又是等比数列.(1)求数列an和bn的通项公式;(2)在cn=1anan+1+(-1)nbn,cn=anbn,cn=2(an+3)anan+1bn+1这三个条件中任选一个,补充在下面问
3、题中,并完成求解.若,求数列cn的前n项和Sn.6.(2021山东菏泽一模)已知等比数列an的前n项和为Sn,且an+1=2Sn+2,数列bn满足b1=2,(n+2)bn=nbn+1.(1)求数列an和bn的通项公式;(2)在an与an+1之间插入n个数,使这n+2个数组成一个公差为cn的等差数列,求数列bncn的前n项和Tn.7.(2021广东广州检测)已知数列an满足a1=1,an+1=3an+3n+1.(1)求证:数列an3n是等差数列;(2)求数列an的通项公式;(3)设数列an的前n项和为Sn,求证:Sn3n3n2-74.专题突破练12求数列的通项及前n项和1.解 (1)设等比数列a
4、n的公比为q(q0),由a32=9a2a6,得a32=9a42,所以q2=19,所以q=13.由2a1+3a2=1,得2a1+3a113=1,所以a1=13.故数列an的通项公式为an=13n.(2)因为bn=log3a1+log3a2+log3an=-(1+2+n)=-n(n+1)2,所以1bn=-2n(n+1)=-21n-1n+1.所以Tn=1b1+1b2+1bn=-21-12+12-13+1n-1n+1=-2nn+1.所以数列1bn的前n项和Tn=-2nn+1.2.解 (1)设an的公差为d,依题意,a1+2d=8,5a1+542d=2(a1+6d),解得a1=2,d=3.所以an=2+
5、3(n-1)=3n-1.(2)因为bn=ancosn+2n+1=(-1)nan+2n+1=(-1)n(3n-1)+2n+1,所以T2n=(a2-a1)+(a4-a3)+(a2n-a2n-1)+(22+23+22n+1)=3n+22(1-22n)1-2=3n+22n+2-4.3.解 (1)由题意可知S5=5(a1+a5)2=5a3=25,所以a3=5.设等差数列an的公差为d,由a3-1,a4+1,a7+3成等比数列,可得(6+d)2=4(8+4d),整理得d2-4d+4=0,解得d=2.所以an=a3+(n-3)d=2n-1.(2)因为bn=(-1)nan+1=(-1)n(2n-1)+1,所以
6、T2n=(-1+1)+(3+1)+(-5+1)+(7+1)+-(4n-3)+1+(4n-1+1)=4n.4.解 (1)设等差数列an的公差为d(d0),则由题意,可知a1+d=3,(a1+d)2=a1(a1+4d),解得a1=1,d=2.an=1+2(n-1)=2n-1.(2)由(1)得an+2=2n+1,Sn=(a1+2)+(a2+2)+(a3+2)+(an-1+2)+(an+2)=3+5+7+(2n-1)+(2n+1)=(2n+1+3)n2=n2+2n.bn=1Sn=1n2+2n=1n(n+2)=121n-1n+2.Tn=b1+b2+b3+bn-1+bn=121-13+12-14+13-1
7、5+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+2=34-2n+32(n+1)(n+2).5.解 (1)设等差数列an的公差为d,等比数列bn的公比为q(q0),由已知得20=b3-a3=a5+b2,即20=3q2-(3+2d),20=(3+4d)+3q,解得d=2,q=3,所以an=2n+1,bn=3n.(2)若选择,则cn=1anan+1+(-1)nbn=1(2n+1)(2n+3)+(-3)n=1212n+1-12n+3+(-3)n,所以Sn=c1+c2+cn=1213-15+(-3)1+1215-17+(-3)2+1212n+1-12n+3+(-3)n=1213-1
8、2n+3+-31-(-3)n1+3=n3(2n+3)-31-(-3)n4.若选择,则cn=anbn=(2n+1)3n,所以Sn=c1+c2+cn=33+532+(2n+1)3n,3Sn=332+533+(2n+1)3n+1,两式相减得-2Sn=32+232+233+23n-(2n+1)3n+1=-2n3n+1,所以Sn=n3n+1.若选择,则cn=2(an+3)anan+1bn+1=2(2n+4)(2n+1)(2n+3)3n+1=1(2n+1)3n-1(2n+3)3n+1,所以Sn=c1+c2+cn=133-1532+1532-1733+1(2n+1)3n-1(2n+3)3n+1=19-1(2
9、n+3)3n+1.6.解 (1)设等比数列an的公比为q,由an+1=2Sn+2,可得an=2Sn-1+2(n2),两式相减得an+1-an=2Sn-2Sn-1=2an,整理得an+1=3an,可知q=3.令n=1,则a2=2a1+2,即3a1=2a1+2,解得a1=2.故an=23n-1.由b1=2,(n+2)bn=nbn+1,得bn+1bn=n+2n,则当n2时,bn=bnbn-1bn-1bn-2b2b1b1=n+1n-1nn-2312=n(n+1).又b1=2满足上式,所以bn=n(n+1).(2)若在an与an+1之间插入n个数,使这n+2个数组成一个公差为cn的等差数列,则an+1-
10、an=(n+1)cn,即23n-23n-1=(n+1)cn,整理得cn=43n-1n+1,所以bncn=4n3n-1,所以Tn=b1c1+b2c2+b3c3+bn-1cn-1+bncn=4130+4231+4332+4(n-1)3n-2+4n3n-1=4130+231+332+(n-1)3n-2+n3n-1,3Tn=4131+232+(n-1)3n-1+n3n,两式相减得-2Tn=4(30+31+32+3n-1-n3n)=41-3n1-3-n3n=(2-4n)3n-2,所以Tn=(2n-1)3n+1.7.(1)证明 由an+1=3an+3n+1,得an+13n+1=an3n+1,即an+13n+1-an3n=1.又a13=13,所以数列an3n是以13为首项,1为公差的等差数列.(2)解 由(1)得an3n=13+(n-1)1=n-23,所以an=n-233n.(3)证明 由(2)得Sn=1-2331+2-2332+(n-1)-233n-1+n-233n,3Sn=1-2332+2-2333+(n-1)-233n+n-233n+1,两式相减得2Sn=n-233n+1-3n+1-92-1=n-763n+1+72,故Sn=n2-7123n+1+74,从而Sn3n=n2-7123n+13n+743n=3n2-74+743n3n2-74.
