2025届高考数学一轮复习专项练习课时规范练28数列的概念

课时规范练28数列的概念基础巩固组1.已知数列5,11,17,A.第19项 B.第20项C.第21项 D.第22项2.记Sn为数列{an}的前n项和.“随意正整数n,均有an>0”是“数列{Sn}是递增数列”的()A.充分不必要条件 B.必要不充分条件C.充要条件 D.既不充分也不必要条件3.(多选)已知数列{an}满足an+1=1-1an(n∈N*),且a1A.a3=-1 B.a2019=1C.S6=3 D.2S2019=20194.在数列{an}中,若a1=1,a2=3,an+2=an+1-an(n∈N*),则该数列的前100项之和是 ()A.18 B.8 C.5 D.25.(多选)已知数列{an}:12,13+23,14+24+34,…,110+210A.an=n2 B.anC.Sn=4nn+1 D.S6.已知{an}是等差数列,且满足:对∀n∈N*,an+an+1=2n,则数列{an}的通项公式an= ()A.n B.n-1C.n-12 D.n+7.已知数列{an}的首项a1=21,且满足(2n-5)an+1=(2n-3)an+4n2-16n+15,则数列{an}的最小的一项是(A.a5 B.a6 C.a7 D.a88.已知每项均大于零的数列{an},首项a1=1且前n项和Sn满足SnSn-1-Sn-1Sn=2SnSn-1(n∈N*A.638 B.639 C.640 D.6419.设Sn为数列{an}的前n项和,且a1=4,an+1=Sn,n∈N*,则S4=.

10.在数列{an}中,a1=2,an+1n+1=ann+ln1+111.已知数列{an}的通项公式为an=n+13n-16(n∈N*),则数列{a12.已知数列{an}满足a1=3,an+1=4an+(1)写出该数列的前4项,并归纳出数列{an}的通项公式;(2)证明:an+1+1综合提升组13.设数列{an}的前n项和为Sn,且a1=1,{Sn+nan}为常数列,则an=()A.13nC.1(n14.已知数列{an}满足an+1-ann=2,a1A.45 B.45-1C.8 D.915.(多选)已知数列{an}的前n项和为Sn(Sn≠0),且满足an+4Sn-1Sn=0(n≥2),a1=14,则下列说法正确的是(A.数列{an}的前n项和为Sn=1B.数列{an}的通项公式为an=1C.数列{an}为递增数列D.数列1Sn为递增数列创新应用组16.已知数列{an}的前n项和为Sn,a1=a,an+1=Sn+3n,若an+1≥an对∀n∈N*成立,则实数a的取值范围是.

17.已知二次函数f(x)=x2-ax+a(a>0,x∈R)有且只有一个零点,数列{an}的前n项和Sn=f(n)(n∈N*).(1)求数列{an}的通项公式;(2)设cn=1-4an(n∈N*),定义全部满足cm·cm+1<0的正整数m的个数,称为这个数列{cn}的变号数,求数列{参考答案课时规范练28数列的概念1.C数列5,11,17,23,29令6n-1=55,得2.A∵an>0,∴数列{Sn}是递增数列,∴“an>0”是“数列{Sn}是递增数列”的充分条件.如数列{an}为-1,1,3,5,7,9,…,明显数列{Sn}是递增数列,但是an不愿定大于零,还有可能小于零,∴数列{Sn}是递增数列不能推出an>0.∴“an>0”是“数列{Sn}是递增数列”的不必要条件.∴“随意正整数n,均有an>0”是“数列{Sn}是递增数列”的充分不必要条件.3.ACD数列{an}满足a1=2,an+1=1-1an(n∈N*),可得a2=12,a3=-1,a4=2,a5=12,…,所以an+3=an,数列的周期为3,a2024=a672×3+3=a3=-4.C∵a1=1,a2=3,an+2=an+1-an∴a3=3-1=2,a4=2-3=-1,a5=-1-2=-3,a6=-3+1=-2,a7=-2+3=1,a8=1+2=3,a9=3-1=2,…∴{an}是周期为6的周期数列,∴S100=S16×6+4=16×(1+3+2-1-3-2)+(1+3+2-1)=5.故选C.5.AC由题意得an=1n+1+2n∴bn=1n2·n+12=4n(n+1)=41n-1n+1,∴数列{bn}的前n项和Sn=b1+b2+b3+…+bn=41-12+12-13+13-16.C由an+an+1=2n,得an+1+an+2=2n+2,两式相减得an+2-an=2=2d,∴d=1,又an+an+d=2n,∴an=n-12.7.A∵4n2-16n+15=(2n-3)(2n-5),∴(2n-5)an+1=(2n-3)an+(2n-3)(2n-5),等式两边同时除以(2n-3)(2n-5),可得an+1可设bn=an2n-5,则b∴bn+1=bn+1,即bn+1-bn=1.∵b1=a12∴数列{bn}是以-7为首项,1为公差的等差数列.∴bn=-7+(n-1)×1=n-8,n∈N*.∴an=(n-8)(2n-5)=2n2-21n+40.可把an看成关于n的二次函数,则依据二次函数的性质,可知其对称轴n=10.52=5∴当n=5时,an取得最小值.故选A.8.C已知SnSn-1-Sn-1Sn=2SnSn-1,数列{an}的每项均大于零,故等号两边同时除以SnSn-1,可得Sn-Sn-1=2,∴{Sn}是以1为首项,2为公差的等差数列,故Sn=2n-1,Sn=(29.32因为Sn为数列{an}的前n项和,且a1=4,an+1=Sn,n∈N*, ①则当n≥2时,an=Sn-1, ②由①-②得an+1-an=an,∴an则数列{an}是从其次项起,公比为2的等比数列,又a2=S1=4,∴an=4·2n-2=2n(n≥2),故an=4所以S4=a5=25=32.10.2n+nlnn由题意得an+1n+1-ann=ln(n+1)-lnn,an∴a22-a11=ln2ann-an-1n-累加得ann-a11=lnn,又a1=2,∴an当n=1时,a1=2,上式成立,故an=2n+nlnn.11.5an=n+13n-16=当n>5时,an>0,且单调递减,当n≤5时,an<0,且单调递减.∴当n=5时,an最小.12.(1)解a1=3,a2=15,a3=63,a4=255.因为a1=41-1,a2=42-1,a3=43-1,a4=44-1,…,所以归纳得an=4n-1.(2)证明因为an+1=4an+3,所以an13.B∵数列{an}的前n项和为Sn,且a1=1,∴S1+1×a1=1+1=2.∵{Sn+nan}为常数列,∴Sn+nan=2.当n≥2时,Sn-1+(n-1)an-1=2,∴(n+1)an=(n-1)an-1,从而a2a1·a3a2·a4a3·…·anan-114.C由an+1-an=2n,知a2-a1=2×1,a3-a2=2×2,…,an-an-1=2(以上各式相加得an-a1=n2-n,n≥2,所以an=n2-n+20,n≥2,当n=1时,a1=20符合上式,所以ann=n+20n-1,n∈所以当n≤4时,ann单调递减,当n≥5时,a因为a44=a55=15.AD由题意,可知数列{an}的前n项和为Sn(Sn≠0),且满足an+4Sn-1Sn=0(n≥2),则Sn-Sn-1=-4Sn-1Sn(n≥2),即1Sn-1S又因为a1=14,所以1S所以数列1Sn是以4为首项,4为公差的等差数列,所以数列1Sn为递增数列,且1Sn=4+(n-1)×4=4n,则Sn又因为当n≥2时,an=Sn-Sn-1=14n-14(所以数列{an}的通项公式为an=1416.[-9,+∞)据题意,得an+1=Sn+1-Sn=Sn+3n,∴Sn+1=2Sn+3n,∴Sn+1-3n+1=2(Sn-3n).又S1-31=a-3,∴数列{Sn-3n}是以a-3为首项,2为公比的等比数列,∴Sn-3n=(a-3)·2n-1即Sn=3n+(a-3)·2n-1.当n=1时,a1=a;当n≥2时,an=Sn-Sn-1=3n+(a-3)×2n-1-3n-1-(a-3)×2n-2=2×3n-1+(a-3)×2n-2,∴an+1-an=4×3n-1+(a-3)×2n-2.又当n≥2时,an+1≥an恒成立,∴a≥3-12×32n-2对∀n∈N*,且n≥2成立,∴又a2=a1+3,∴a2≥a1成立.综上,所求实数a的取值范围是[-9,+∞).17.解(1)依题意,得Δ=a2-4a=0,所以a=0或a=4.又由a>0得a=4,