专题突破练13等差、等比数列的综合问题1.已知Sn是等差数列{an}的前n项和,且a3=-6,S5=S6.(1)求{an}的通项公式;(2)若等比数列{bn}满足b1=a2,b2=S3,求{bn}的前n项和.2.(2019北京丰台区高三第二学期综合练习二)已知数列{an}满足a1=1,an+1=e·an(e是自然对数的底数,n∈N*).(1)求{an}的通项公式;(2)设数列{lnan}的前n项和为Tn,求证:当n≥2时,1T2+1T3+…+1Tn<2.3.设等差数列{an}的公差不为0,a2=1,且a2,a3,a6成等比数列.(1)求{an}的通项公式;(2)设数列{an}的前n项和为Sn,求使Sn>35成立的n的最小值.4.已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.5.(2019河北保定高三第二次模拟考试)已知函数f(x)=log3(ax+b)的图象经过点A(2,1)和B(5,2),an=an+b,n∈N*.(1)求an;(2)设数列{an}的前n项和为Sn,bn=2n+2❑√Sn,求{bn}的前n项和Tn.6.(2019广东佛山第一中学高三上学期期中考试)等差数列{an}中,a1=3,前n项和为Sn,等比数列{bn}各项均为正数,b1=1,且b2+S2=12,{bn}的公比q=S2b2.(1)求an与bn;(2)求Tn=Sb1+Sb2+Sb3+Sb4+…+Sbn.7.已知{an}是首项为1的等比数列,数列{bn}满足b1=2,b2=5,且anbn+1=anbn+an+1.(1)求数列{an}的通项公式;(2)求数列{bn}的前n项和.8.(2019黑龙江哈尔滨第三中学高三上学期期中考试)已知数列{an}中,a1=32且an=12(an-1+n+1)(n≥2,n∈N*).(1)求a2,a3,并证明{an-n}是等比数列;(2)设bn=2n·an,求数列{bn}的前n项和Sn.参考答案专题突破练13等差、等比数列的综合问题1.解(1)设等差数列{an}的公差为d.因为S5=S6,所以a6=a3+3d=0.因为a3=-6,所以d=2,a1=-10.所以an=2n-12.(2)设等比数列{bn}的公比为q.由(1)可知,b1=-8,b2=-24,所以q=3.故数列{bn}的前n项和为-8(1-3n)1-3=4(1-3n).2.(1)解因为a1=1,an+1=e·an(n∈N*),所以数列{an}是以1为首项,e为公比的等比数列,所以an=en-1.(2)证明由(1)知,lnan=lnen-1=n-1,所以Tn=0+1+2+…+(n-1)=n(n-1)2,所以1T2+1T3+…+1Tn=21×2+22×3+23×4+…+2n(n-1)=21-12+12−13+13−14+…+1n-1−1n=21-1n.因为1n>0,所以1-1n<1.所以21-1n<2.即1T2+1T3+…+1Tn<2.3.解(1)设等差数列{an}的公差为d,d≠0.∵a2,a3,a6成等比数列,∴a32=a2·a6,即(1+d)2=1+4d,解得d=2或d=0(舍去d=0),∴an=a2+(n-2)d=2n-3.(2)∵an=2n-3,∴Sn=n(a1+an)2=n(a2+an-1)2=n2-2n.依题意有n2-2n>35,解得n>7.故使Sn>35成立的n的最小值为8.4.解(1)∵3S1,2S2,S3成等差数列,∴4S2=3S1+S3,∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,∴an=a1qn-1=3n.(2)由(1)知,bn=log3an=log33n=n,∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,∴Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)=-4(1+2+…+n)=-4×n(n+1)2=-2n2-2n.5.解(1)由函数f(x)=log3(ax+b)的图象经过点A(2,1)和B(5,2),得{log3(2a+b)=1,log3(5a+b)=2,解得{a=2,b=-1.所以an=2n-1,n∈N*.(2)由(1)知数列{an}为以1为首项,2为公差的等差数列,所以Sn=n+n(n-1)2×2=n2,得bn=2n+2❑√Sn=2n+2n.∴Tn=(2×1+21)+(2×2+22)+(2×3+23)+…+(2×n+2n)=2×(1+2+3+…+n)+(21+22+23+…+2n)=2×(1+n)n2+2(2n-1)2-1=2n+1+n2+n-2.6.解(1)由已知可得{q+3+a2=12,q=3+a2q,解得q=3或q=-4(舍去负值),a2=6.∴an=3n,bn=3n-1.(2)∵Sn=3n(n+1)2,∴Sbn=32bn(bn+1)=32¿+bn),∴Tn=Sb1+Sb2+Sb3+Sb4+…+Sbn=32(b1+b2+…+bn)+(b12+b22+…+bn2)=32(30+31+…+3n-1)+(30+32+…+32n-2)==321-3n1-3+1-32n1-32=32n+116+3n+14−1516.7.解(1)把n=1代入已知等式得a1b2=a1b1+a2,∴a2=a1b2-a1b1=3a1.∴{an}是首项为1,公比为3的等比数列,即an=3n-1.(2)由已知得bn+1-bn=an+1an=3,∴{bn}是首项为2,公差为3的等差数列,其通项公式为bn=3n-1,∴Sn=n(b1+bn)2=n(2+3n-1)2=3n2+n2.8.解(1)由题意,可知:a2=12(a1+2+1)=1232+2+1=94,a3=12(a2+3+1)=1294+3+1=258.①当n=1时,a1-1=32-1=12,②当n≥2时,an-n=12(an-1+n+1)-n=12an-1+12n+12-n=12an-1-12n+12=12(an-1-n+1)=12[an-1-(n-1)].∴数列{an-n}是以12为首项,12为公比的等比数列.(2)由(1)可知,an-n=12n,∴an=n+12n,n∈N*.∴bn=2n·an=2n·n+12n=n·2n+2n·12n=n·2n+1.∴Sn=b1+b2+b3+…+bn=(1·21+1)+(2·22+1)+(3·23+1)+…+(n·2n+1),∴Sn=1·21+2·22+3·23+…+n·2n+n,③2Sn=1·22+2·23+…+(n-1)·2n+n·2n+1+2n,④由③-④,可得:-Sn=1·21+1·22+1·23+…+1·2n-n·2n+1+n-2n=2-2n+11-2-n·2n+1-n=(1-n)·2n+1-n-2,∴Sn=(n-1)·2n+1+n+2.
