2025年高考數(shù)學(xué)一輪復(fù)知識(shí)點(diǎn)復(fù)習(xí)-6.4數(shù)列求和-專項(xiàng)訓(xùn)練【含解析】

PAGE6.4數(shù)列求和-專項(xiàng)訓(xùn)練【原卷版】(時(shí)間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)若數(shù)列{an}的通項(xiàng)公式為an=2n+2n-1,則該數(shù)列的前10項(xiàng)和為()A.2146 B.1122C.2148 D.11242.(5分)在等差數(shù)列{an}中,若a3+a5+a7=6,a11=8,則數(shù)列{1an+3an+4}的前nA.n+1n+2 C.nn+2 D3.(5分)數(shù)列{(-1)n(2n-1)}的前2024項(xiàng)和S2024等于()A.-2022 B.2022C.-2024 D.20244.(5分)已知數(shù)列{an}滿足a1=1,且對(duì)任意的n∈N*都有an+1=a1+an+n,則1an的前100項(xiàng)和為(A.100101 B.99C.101100 D.5.(5分)122-1+132-A.n+12(n+2)C.34-12(1n+1+1n+2) D6.(5分)(多選題)已知數(shù)列{an}:12,13+23,14+24+34,…,110+210+…+910,…,若bn=1anA.an=n2 B.an=C.Sn=4nn+1 D.S7.(5分)數(shù)列{an}的通項(xiàng)公式為an=1n+n+1,若{an}的前n項(xiàng)和為24,則8.(5分)已知數(shù)列{an}的首項(xiàng)為-1,anan+1=-2n,則數(shù)列{an}的前10項(xiàng)之和S10=.

9.(5分)已知數(shù)列{an}滿足an+1=nn+1an,a1=1,則數(shù)列{anan+1}的前10項(xiàng)和為10.(10分)(2023·西安模擬)已知數(shù)列an滿足a1=1,an+1-an=n+1(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列1an的前n項(xiàng)和為Sn,求數(shù)列l(wèi)og2Sn11.(10分)(2023·惠州模擬)記Sn是公差不為零的等差數(shù)列an的前n項(xiàng)和,若S3=6,a3是a1和a9的等比中項(xiàng)(1)求數(shù)列an(2)記bn=1an·a【加練備選】(2023·廣州模擬)已知等差數(shù)列an的前n項(xiàng)和Sn滿足S3=-3,S7=-21(1)求an(2)bn=-an+1,求數(shù)列1bnbn+2的前【能力提升練】12.(5分)(2021·新高考Ⅰ卷)某校學(xué)生在研究民間剪紙藝術(shù)時(shí),發(fā)現(xiàn)剪紙時(shí)經(jīng)常會(huì)沿紙的某條對(duì)稱軸把紙對(duì)折.規(guī)格為20dm×12dm的長(zhǎng)方形紙,對(duì)折1次共可以得到10dm×12dm,20dm×6dm兩種規(guī)格的圖形,它們的面積之和S1=240dm2,對(duì)折2次共可以得到5dm×12dm,10dm×6dm,20dm×3dm三種規(guī)格的圖形,它們的面積之和S2=180dm2,以此類推,則對(duì)折4次共可以得到不同規(guī)格圖形的種數(shù)為;如果對(duì)折n次,那么∑k=1nSk=13.(5分)已知數(shù)列{an}和{bn}滿足a1a2a3·…·an=2bn(n∈N*).若數(shù)列{an}為等比數(shù)列,且a1=2,a4=16,則數(shù)列{bn}的通項(xiàng)公式bn=,數(shù)列{1bn}的前n項(xiàng)和S14.(10分)已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,an+1=(λ+1)Sn+1(n∈N*,λ≠-2)且3a1,4a2,a3+13成等差數(shù)列(1)求數(shù)列an(2)若anbn=log4an+1,求數(shù)列bn的前n項(xiàng)和Tn6.4數(shù)列求和-專項(xiàng)訓(xùn)練【解析版】(時(shí)間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)若數(shù)列{an}的通項(xiàng)公式為an=2n+2n-1,則該數(shù)列的前10項(xiàng)和為()A.2146 B.1122C.2148 D.1124【解析】選A.因?yàn)閍n=2n+2n-1,所以前n項(xiàng)和Sn=2(1-2n)1-2所以前10項(xiàng)和S10=211+102-2=2146.2.(5分)在等差數(shù)列{an}中,若a3+a5+a7=6,a11=8,則數(shù)列{1an+3an+4}的前nA.n+1n+2 C.nn+2 D【解析】選B.設(shè)等差數(shù)列{an}的公差為d,由a3+a5+a7=6,a11=8,得a5=2,d=1,所以an=n-3,則an+3=n,an+4=n+1,所以1an+3an+4=1n(n3.(5分)數(shù)列{(-1)n(2n-1)}的前2024項(xiàng)和S2024等于()A.-2022 B.2022C.-2024 D.2024【解析】選D.S2024=-1+3-5+7-…-(2×2023-1)+(2×2024-1)=2×1012=2024.4.(5分)已知數(shù)列{an}滿足a1=1,且對(duì)任意的n∈N*都有an+1=a1+an+n,則1an的前100項(xiàng)和為(A.100101 B.99C.101100 D.【解析】選D.因?yàn)閍n+1=a1+an+n,a1=1,所以an+1-an=1+n,所以an-an-1=n(n≥2),所以an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n(所以1an=2n(n所以1an的前100項(xiàng)和為2(1-12+12-13+…+1100-5.(5分)122-1+132-A.n+12(n+2)C.34-12(1n+1+1n+2) D【解題指導(dǎo)】先化簡(jiǎn)通項(xiàng)公式,再裂項(xiàng)求和.【解析】選C.因?yàn)?(n+1)2-1=1n2所以122-1+1=12(1-13+12-14+13-15+…+1n-1n+2)=12(32-1n6.(5分)(多選題)已知數(shù)列{an}:12,13+23,14+24+34,…,110+210+…+910,…,若bn=1anA.an=n2 B.an=C.Sn=4nn+1 D.S【解析】選AC.由題意得an=1n+1+2n+1+…+nn所以bn=1n2·n+12=所以數(shù)列{bn}的前n項(xiàng)和Sn=b1+b2+b3+…+bn=4(1-12+12-13+13-14=4(1-1n+1)=7.(5分)數(shù)列{an}的通項(xiàng)公式為an=1n+n+1,若{an}的前n項(xiàng)和為24,則【解析】an=n+1-n,所以Sn=(2-1)+(3-2)+…+(n+1-n)=n+1-1,令Sn=24,得答案:6248.(5分)已知數(shù)列{an}的首項(xiàng)為-1,anan+1=-2n,則數(shù)列{an}的前10項(xiàng)之和S10=.

【解析】因?yàn)閍nan+1=-2n,所以an+1an+2=-2n+1,兩式相除可得an所以{an}的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)均為公比為2的等比數(shù)列,又a1=-1,a2=-2所以S10=(a1+a3+…+a9)+(a2+a4+…+a10)=-1×(1-答案:319.(5分)已知數(shù)列{an}滿足an+1=nn+1an,a1=1,則數(shù)列{anan+1}的前10項(xiàng)和為【解析】因?yàn)閍n+1=nn+1an,a所以(n+1)·an+1=nan,所以數(shù)列{nan}是每項(xiàng)均為1的常數(shù)列,所以nan=1,所以an=1n,anan+1=1n(n+1所以數(shù)列{anan+1}的前10項(xiàng)和為(11-12)+(12-13)+…+(110-1答案:1010.(10分)(2023·西安模擬)已知數(shù)列an滿足a1=1,an+1-an=n+1(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列1an的前n項(xiàng)和為Sn,求數(shù)列l(wèi)og2Sn【解析】(1)由題意數(shù)列an滿足a1=1,an+1-an=n則an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n((2)由(1)可得1an=2(1n故Sn=2(1-12+12-13+…+1n-所以log2Sn=log22nn+1=1+log故Tn=n+log2(12×23×34×…×nn+1)=n+log21n+1=11.(10分)(2023·惠州模擬)記Sn是公差不為零的等差數(shù)列an的前n項(xiàng)和,若S3=6,a3是a1和a9的等比中項(xiàng)(1)求數(shù)列an(2)記bn=1an·a【解析】(1)由題意知a32=a1·a9,設(shè)等差數(shù)列an的公差為d,則a1(a1+8d因?yàn)閐≠0,解得a1=d,又S3=3a1+3d=6,可得a1=d=1,所以數(shù)列an是首項(xiàng)為1和公差為1的等差數(shù)列,所以an=a1+(n-1)d=n,n∈N*(2)由(1)可知bn=1n(n+1)(n+2)設(shè)數(shù)列bn的前n項(xiàng)和為TnTn=12[11×2-12×3+12×3-13×4+…+1n(n+1所以T20=12×(12-121×22所以數(shù)列bn的前20項(xiàng)和為115【加練備選】(2023·廣州模擬)已知等差數(shù)列an的前n項(xiàng)和Sn滿足S3=-3,S7=-21(1)求an(2)bn=-an+1,求數(shù)列1bnbn+2的前【解析】(1)設(shè)公差為d,則S3=3a1+3d=-3,S7=7a1+21d=-21,所以3a1+3所以an=a1+(n-1)d=-n+1;(2)bn=n,所以1bnbn+2=1n(所以Tn=12(1-13)+12(12-14)+12(13-15)+…+12(1=12(1+12-1n+1-1n【能力提升練】12.(5分)(2021·新高考Ⅰ卷)某校學(xué)生在研究民間剪紙藝術(shù)時(shí),發(fā)現(xiàn)剪紙時(shí)經(jīng)常會(huì)沿紙的某條對(duì)稱軸把紙對(duì)折.規(guī)格為20dm×12dm的長(zhǎng)方形紙,對(duì)折1次共可以得到10dm×12dm,20dm×6dm兩種規(guī)格的圖形,它們的面積之和S1=240dm2,對(duì)折2次共可以得到5dm×12dm,10dm×6dm,20dm×3dm三種規(guī)格的圖形,它們的面積之和S2=180dm2,以此類推,則對(duì)折4次共可以得到不同規(guī)格圖形的種數(shù)為;如果對(duì)折n次,那么∑k=1nSk=【解析】依題意得,S1=120×2=240(dm2);S2=60×3=180(dm2);當(dāng)n=3時(shí),共可以得到5dm×6dm,52dm×12dm,10dm×3dm,20dm×32dm四種規(guī)格的圖形,且5×6=30,52所以S3=30×4=120(dm2);當(dāng)n=4時(shí),共可以得到5dm×3dm,5254dm×12dm,10dm×32dm,20dm×34dm五種規(guī)格的圖形,所以對(duì)折4次共可以得到不同規(guī)格圖形的種數(shù)為5,且5×3=15,52×6=15,54×12=15,10×32=15,20×3……所以可歸納Sk=2402k·(k+1)=240(k所以∑k=1nSk=240(1+322+42所以12×∑k=1nSk=240×(222+323由①-②得,12·∑k=1nSk=240(1+122+123+124+…+12所以∑k=1nSk=240(3-n答案:5240×(3-n+313.(5分)已知數(shù)列{an}和{bn}滿足a1a2a3·…·an=2bn(n∈N*).若數(shù)列{an}為等比數(shù)列,且a1=2,a4=16,則數(shù)列{bn}的通項(xiàng)公式bn=,數(shù)列{1bn}的前n項(xiàng)和S【解析】因?yàn)閿?shù)列{an}為等比數(shù)列,且a1=2,a4=16,所以公比q=3a4a所以an=2n,所以a1a2a3·…·an=21×22×23×…×2n=21+2+3+…+n=2n因?yàn)閍1a2a3·…·an=2bn,所以bn=所以1bn=2n(n所以數(shù)列{1bn}的前Sn=1b1+1b2+1b3+…+1bn=2(11-12+12-13+答案:n(n14.(10分)已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,an+1=(λ+1)Sn+1(n∈N*,λ≠-2)且3a1,4a2,a3+13成等差數(shù)列(1)求數(shù)列an(2)若anbn=log4an+1,求數(shù)列bn的前n項(xiàng)和Tn【解析】(1)因?yàn)閍n+1=(λ+1)Sn+1,a1=1,當(dāng)n=1時(shí),a2=(λ+1)S1+1=λ+2,當(dāng)n≥2時(shí),an=(λ+1)Sn-1+1,所以an+1