2025版新教材高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練28數(shù)列的概念含解析新人教A版

課時(shí)規(guī)范練28數(shù)列的概念基礎(chǔ)鞏固組1.若數(shù)列的前4項(xiàng)分別是12,-13,14,-1A.(-1)nC.(-1)n2.記Sn為數(shù)列{an}的前n項(xiàng)和.“隨意正整數(shù)n,均有an>0”是“數(shù)列{Sn}是遞增數(shù)列”的()A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件3.(多選)已知數(shù)列{an}滿意an+1=1-1an(n∈N*),且a1=2,A.a3=-1 B.a2019=1C.S6=3 D.2S2019=20194.(2024河北保定高三期末)在數(shù)列{an}中,若a1=1,a2=3,an+2=an+1-an(n∈N*),則該數(shù)列的前100項(xiàng)之和是()A.18 B.8 C.5 D.25.(多選)已知數(shù)列{an}:12,13+23,14+24+34,…,110+210+310+…+910A.an=n2 B.anC.Sn=4nn+1 D.S6.(2024湖南益陽(yáng)高三期末)已知{an}是等差數(shù)列,且滿意:對(duì)?n∈N*,an+an+1=2n,則數(shù)列{an}的通項(xiàng)公式an=()A.n B.n-1C.n-12 D.n+7.已知數(shù)列{an}的首項(xiàng)a1=21,且滿意(2n-5)an+1=(2n-3)an+4n2-16n+15,則數(shù)列{an}的最小的一項(xiàng)是(A.a5 B.a6 C.a7 D.a88.已知每項(xiàng)均大于零的數(shù)列{an},首項(xiàng)a1=1且前n項(xiàng)和Sn滿意SnSn-1-Sn-1Sn=2SnSn-1(n∈N*且nA.638 B.639 C.640 D.6419.設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,且a1=4,an+1=Sn,n∈N*,則S4=.

10.在數(shù)列{an}中,a1=2,an+1n+1=ann+ln1+1n,11.已知數(shù)列{an}的通項(xiàng)公式為an=n+13n-16(n∈N*),則數(shù)列{an}12.已知數(shù)列{an}滿意a1=3,an+1=4an+(1)寫(xiě)出該數(shù)列的前4項(xiàng),并歸納出數(shù)列{an}的通項(xiàng)公式;(2)證明:an+1+1綜合提升組13.(2024廣東中山期末)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且a1=1,{Sn+nan}為常數(shù)列,則an=()A.13n-C.1(n+114.(2024安徽江淮十校第三次聯(lián)考)已知數(shù)列{an}滿意an+1-ann=2,a1=20,A.45 B.45-1C.8 D.915.(多選)(2024江西贛州教化發(fā)展聯(lián)盟2月聯(lián)考)已知數(shù)列{an}的前n項(xiàng)和為Sn(Sn≠0),且滿意an+4Sn-1Sn=0(n≥2),a1=14,則下列說(shuō)法正確的是(A.數(shù)列{an}的前n項(xiàng)和為Sn=1B.數(shù)列{an}的通項(xiàng)公式為an=1C.數(shù)列{an}為遞增數(shù)列D.數(shù)列1Sn為遞增數(shù)列創(chuàng)新應(yīng)用組16.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=a,an+1=Sn+3n,若an+1≥an對(duì)?n∈N*成立,則實(shí)數(shù)a的取值范圍是.

17.已知二次函數(shù)f(x)=x2-ax+a(a>0,x∈R)有且只有一個(gè)零點(diǎn),數(shù)列{an}的前n項(xiàng)和Sn=f(n)(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)cn=1-4an(n∈N*),定義全部滿意cm·cm+1<0的正整數(shù)m的個(gè)數(shù),稱為這個(gè)數(shù)列{cn}的變號(hào)數(shù),求數(shù)列{c參考答案課時(shí)規(guī)范練28數(shù)列的概念1.A由于數(shù)列的前4項(xiàng)分別是12,-13,14,-15,可得奇數(shù)項(xiàng)為正數(shù),偶數(shù)項(xiàng)為負(fù)數(shù),第n項(xiàng)的肯定值等于1n+12.A∵an>0,∴數(shù)列{Sn}是遞增數(shù)列,∴“an>0”是“數(shù)列{Sn}是遞增數(shù)列”的充分條件.如數(shù)列{an}為-1,1,3,5,7,9,…,明顯數(shù)列{Sn}是遞增數(shù)列,但是an不肯定大于零,還有可能小于零,∴數(shù)列{Sn}是遞增數(shù)列不能推出an>0.∴“an>0”是“數(shù)列{Sn}是遞增數(shù)列”的不必要條件.∴“隨意正整數(shù)n,均有an>0”是“數(shù)列{Sn}是遞增數(shù)列”的充分不必要條件.3.ACD數(shù)列{an}滿意a1=2,an+1=1-1an(n∈N*),可得a2=12,a3=-1,a4=2,a5=12,…,所以an+3=an,數(shù)列的周期為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}是周期為6的周期數(shù)列,∴S100=S16×6+4=16×(1+3+2-1-3-2)+(1+3+2-1)=5.故選C.5.AC由題意得an=1n+1+2n∴bn=1n2·n+12=4n(n+1)=41n-1n+1,∴數(shù)列{bn}的前n項(xiàng)和Sn=b1+b2+b3+…+bn=41-12+12-13+13-14+6.C由an+an+1=2n,得an+1+an+2=2n+2,兩式相減得an+2-an=2=2d,∴d=1,又an+an+d=2n,∴an=n-12.故選7.A∵4n2-16n+15=(2n-3)(2n-5),∴(2n-5)an+1=(2n-3)an+(2n-3)(2n-5),等式兩邊同時(shí)除以(2n-3)(2n-5),可得an+1可設(shè)bn=an2n-5,則b∴bn+1=bn+1,即bn+1-bn=1.∵b1=a12∴數(shù)列{bn}是以-7為首項(xiàng),1為公差的等差數(shù)列.∴bn=-7+(n-1)×1=n-8,n∈N*.∴an=(n-8)(2n-5)=2n2-21n+40.可把a(bǔ)n看成關(guān)于n的二次函數(shù),則依據(jù)二次函數(shù)的性質(zhì),可知其對(duì)稱軸n=10.52=5∴當(dāng)n=5時(shí),an取得最小值.故選A.8.C已知SnSn-1-Sn-1Sn=2SnSn-1,數(shù)列{an}的每項(xiàng)均大于零,故等號(hào)兩邊同時(shí)除以SnSn-1,可得Sn-Sn-1=2,∴{Sn}是以1為首項(xiàng),2為公差的等差數(shù)列,故Sn=2n-1,Sn=(2n-9.32因?yàn)镾n為數(shù)列{an}的前n項(xiàng)和,且a1=4,an+1=Sn,n∈N*,①則當(dāng)n≥2時(shí),an=Sn-1,②由①-②得an+1-an=an,∴an則數(shù)列{an}是從其次項(xiàng)起,公比為2的等比數(shù)列,又a2=S1=4,∴an=4·2n-2=2n(n≥2),故an=4所以S4=a5=25=32.10.2n+nlnn由題意得an+1n+1-ann=ln(n+1)-lnn,ann∴a22-a11=ln2-ln1,ann-an-1n-1累加得ann-a11=lnn,又a1=2,∴ann當(dāng)n=1時(shí),a1=2,上式成立,故an=2n+nlnn.11.5an=n+13n-16=當(dāng)n>5時(shí),an>0,且單調(diào)遞減,當(dāng)n≤5時(shí),an<0,且單調(diào)遞減.∴當(dāng)n=5時(shí),an最小.12.(1)解a1=3,a2=15,a3=63,a4=255.因?yàn)閍1=41-1,a2=42-1,a3=43-1,a4=44-1,…,所以歸納得an=4n-1.(2)證明因?yàn)閍n+1=4an+3,所以an13.B∵數(shù)列{an}的前n項(xiàng)和為Sn,且a1=1,∴S1+1×a1=1+1=2.∵{Sn+nan}為常數(shù)列,∴Sn+nan=2.當(dāng)n≥2時(shí),Sn-1+(n-1)an-1=2,∴(n+1)an=(n-1)an-1,從而a2a1·a3a2·a4a3·…·anan-1=13·214.C由an+1-an=2n,知a2-a1=2×1,a3-a2=2×2,…,an-an-1=2(n-以上各式相加得an-a1=n2-n,n≥2,所以an=n2-n+20,n≥2,當(dāng)n=1時(shí),a1=20符合上式,所以ann=n+20n-1,n∈所以當(dāng)n≤4時(shí),ann單調(diào)遞減,當(dāng)n≥5時(shí),a因?yàn)閍44=a55=8,所以a15.AD由題意,可知數(shù)列{an}的前n項(xiàng)和為Sn(Sn≠0),且滿意an+4Sn-1Sn=0(n≥2),則Sn-Sn-1=-4Sn-1Sn(n≥2),即1Sn-1Sn又因?yàn)閍1=14,所以1S所以數(shù)列1Sn是以4為首項(xiàng),4為公差的等差數(shù)列,所以數(shù)列1Sn為遞增數(shù)列,且1Sn=4+(n-1)×4=4n,則Sn又因?yàn)楫?dāng)n≥2時(shí),an=Sn-Sn-1=14n-14(所以數(shù)列{an}的通項(xiàng)公式為an=14,n16.[-9,+∞)據(jù)題意,得an+1=Sn+1-Sn=Sn+3n,∴Sn+1=2Sn+3n,∴Sn+1-3n+1=2(Sn-3n).又S1-31=a-3,∴數(shù)列{Sn-3n}是以a-3為首項(xiàng),2為公比的等比數(shù)列,∴Sn-3n=(a-3)·2n-1即Sn=3n+(a-3)·2n-1.當(dāng)n=1時(shí),a1=a;當(dāng)n≥2時(shí),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.又當(dāng)n≥2時(shí),an+1≥an恒成立,∴a≥3-12×32n-2對(duì)?n∈N*,且n≥2成立,∴又a2=a1+3,∴a2≥a1成立.綜上,所求實(shí)數(shù)a的取值范圍是[-9,+∞).17.解(1)依題意,得Δ=a2-4a=0,所以a=0或a=4.又由