備戰(zhàn)2025年高考二輪復(fù)習(xí)數(shù)學(xué)專題突破練11　數(shù)列通項(xiàng)與求和(提升篇)

專題突破練(分值:91分)學(xué)生用書P163主干知識(shí)達(dá)標(biāo)練1.(15分)(2024河北唐山一模)已知數(shù)列{an}是正項(xiàng)等比數(shù)列,其前n項(xiàng)和為Sn,且a2a4=16,S5=S3+24.(1)求{an}的通項(xiàng)公式;(2)記{an+log2an}的前n項(xiàng)和為Tn,求滿足Tn<2024的最大整數(shù)n.解(1)設(shè){an}的公比為q(q>0),則an=a1qn-1,依題意可得a整理得q2+q-6=0,解得q=2或q=-3(舍去),所以an=a3qn-3=2n-1.(2)由(1)可知an+log2an=2n-1+n-1,故Tn=(20+21+22+…+2n-1)+(0+1+2+…+n-1)=2n-1+n(顯然,Tn隨著n的增大而增大,T10=210-1+45=1068<2024,T11=211-1+55=2102>2024,所以滿足Tn<2024的最大整數(shù)n=10.2.(10分)(2022新高考Ⅰ,17)記Sn為數(shù)列{an}的前n項(xiàng)和,已知a1=1,Snan是公差為(1)求{an}的通項(xiàng)公式;(2)證明:1a1+1a2+(1)解(方法一)∵Snan是以S1a1=1∴Snan=1+(n-∴Sn=n+23an.當(dāng)n≥2時(shí),Sn-1=n+13an-1.①-②得an=Sn-Sn-1=n+23an-n+13∴n+13an-1=n-13a∴an=anan-1·an-1an-2·…·a2又a1=1,∴an=(n+1)×n2×1×1又當(dāng)n=1時(shí),a1=1也符合上式,∴an=n((方法二)∵Snan是以S1a1=1為首項(xiàng),以13為公差的等差數(shù)列,∴Sn∴Sn=n+23an.當(dāng)n≥2時(shí),Sn-1=n+13an-1.①-②得an=Sn-Sn-1=n+23an-n+13an-1,∴n+13an-∴ann+1=設(shè)ann(n+1)=bn,則∴{bn}為常數(shù)列,且b1=a1∴ann(n+1)=bn=12(方法三)∵Snan是首項(xiàng)為S1a1=1,公差為13的等差數(shù)列,∴Snan=1+13(n-∴SnSn∴Sn=S1·S2S1·S3S2·…·∴an=Sn-Sn-1=n(n+1又a1=1滿足此公式,∴{an}的通項(xiàng)公式為an=n((2)證明由(1)知,1an=2∴1a1+1a2+…+1an=21-12+12-13+3.(15分)(2024浙江金華高三期末)已知數(shù)列{an}是等差數(shù)列,a1=3,d≠0,且a1,a7,a25構(gòu)成等比數(shù)列.(1)求an;(2)設(shè)f(n)=an,若存在數(shù)列{bn}滿足b1=1,b2=7,b3=25,且數(shù)列{f(bn)}為等比數(shù)列,求{anbn}的前n項(xiàng)和Sn.解(1)∵{an}是等差數(shù)列,a1=3,d≠0,∴a7=a1+6d,a25=a1+24d.∵a1,a7,a25構(gòu)成等比數(shù)列,∴(a1+6d)2=a1(化簡(jiǎn)可得a1=3d=3,∴d=1,∴an=n+2.(2)∵f(b1)=f(1)=a1=3,f(b2)=f(7)=a7=9,f(b3)=f(25)=a25=27,又?jǐn)?shù)列{f(bn)}為等比數(shù)列,∴首項(xiàng)為3,公比為3,故f(bn)=3n,而f(bn)=abn=bn∴3n=bn+2,∴bn=3n-2,∴anbn=(n+2)3n-2(n+2),設(shè)數(shù)列{(n+2)3n}的前n項(xiàng)和為Tn,則Tn=(1+2)×31+(2+2)×32+…+(n+2)×3n①,3Tn=(1+2)×32+(2+2)×33+…+(n+2)×3n+1②,①②相減得-2Tn=(1+2)×31+32+…+3n-(n+2)×3n+1,化簡(jiǎn)可得Tn=-129+9×(1-3n-1)1-又因?yàn)榈炔顢?shù)列{2(n+2)}的前n項(xiàng)和為n(2×3+2n+4)2綜上可得Sn=(2n+3)×3n+1-關(guān)鍵能力提升練4.(15分)(2024山東德州高三開學(xué)考試)已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足6Sn=(3n+2)an+2.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=(-1)n+1(6n+1)ana解(1)因?yàn)?Sn=(3n+2)an+2,當(dāng)n=1時(shí),6S1=6a1=5a1+2,所以a1=2,當(dāng)n≥2時(shí),6Sn-1=(3n-1)an-1+2,所以6Sn-6Sn-1=6an=(3n+2)an-(3n-1)an-1,所以anan-1累乘得anan-1·an-所以an=3n-1(n≥2),當(dāng)n=1時(shí)a1=2也符合上式,所以an=3n-1.(2)由(1)得bn=(-1)n+1(6n+1)a所以T100=12+15-5.(15分)(2024湖南永州模擬)已知數(shù)列{an}的前n項(xiàng)和為Sn,對(duì)于任意的n∈N*,都有點(diǎn)P(an,Sn)在直線2x-3y+1=0上.(1)求數(shù)列{an}的通項(xiàng)公式;(2)記數(shù)列{an}的前n項(xiàng)中的最大值為Mn,最小值為mn,令bn=Mn+mn2,求數(shù)列{bn}的前解(1)根據(jù)題意得,對(duì)于任意的n∈N*都有3Sn=2an+1,當(dāng)n≥2時(shí),3Sn-1=2an-1+1,兩式相減得3(Sn-Sn-1)=(2an+1)-(2an-1+1),即3an=2an-2an-1(n≥2),整理得an=-2an-1(n≥2),當(dāng)n=1時(shí),3S1=2a1+1,故a1=1,所以數(shù)列{an}是首項(xiàng)為1,公比為-2的等比數(shù)列,所以an=(-2)n-1.(2)當(dāng)n為奇數(shù)時(shí),an=2n-1,且an>0,當(dāng)n為偶數(shù)時(shí),an=-2n-1,且an<0,因此當(dāng)n為大于1的奇數(shù)時(shí),{an}前n項(xiàng)中的最大值為an=(-2)n-1,最小值為an-1=(-2)n-2,此時(shí)bn=Mn當(dāng)n為偶數(shù)時(shí),{an}的前n項(xiàng)中的最大值為an-1=(-2)n-2,最小值為an=(-2)n-1,此時(shí)bn=Mn當(dāng)n=1時(shí),b1=a1,因此T20=b1+(b3+b5+…+b19)+(b2+b4+b6+…+b20)=a1+a3+a22+a5=a12+S=1-核心素養(yǎng)創(chuàng)新練6.(多選題)(2024山東日照高三期末)在平面四邊形ABCD中,點(diǎn)D為動(dòng)點(diǎn),△ABD的面積是△BCD面積的3倍,又?jǐn)?shù)列{an}滿足a1=3,恒有BD=(an-3n-1)BA+(an+1+3n)BC,設(shè){an}的前n項(xiàng)和為Sn,則()A.{an}為等比數(shù)列B.a4=-81C.an3nD.Sn=(3-n)3n-3答案BCD解析設(shè)AC,BD交于E點(diǎn),則S△ABD即AE=3EC,故BE=BA+AE=BA+34AC=BA+34(BC-BA)=14BA+34BC,由于B,E,D三點(diǎn)共線,故存在實(shí)數(shù)λ(λ≠0,1),使得BD=λBE,即得BD=(an-3n-1)BA+(an+即an+1=3an-2×3n,則an+13n+1=而a1=3,故an3n為首項(xiàng)是a13=1,公差為-23則an3n=1+(n-1)×-23=5-2n3,故a故a4=5-2×43×34=-又an+1an=5-2(n+1)3·3nSn=a1+a2+…+an=33×3+13×32+-13×33+…+故3Sn=33×32+13×33+-13×34+…+7-2n3兩式相減得-2Sn=3-23×(32+33+34+…+3n)-5-2n3·3n+1=3-23×9(1-3n-1)1-3-5-2n3·3n+17.(15分)(2024河北滄州一模)在數(shù)列{an}中,已知a1+a22+a322(1)求數(shù)列{an}的通項(xiàng)公式;(2)在數(shù)列{an}中的a1和a2之間插入1個(gè)數(shù)x11,使a1,x11,a2成等差數(shù)列;在a2和a3之間插入2個(gè)數(shù)x21,x22,使a2,x21,x22,a3成等差數(shù)列;…;在an和an+1之間插入n個(gè)數(shù)xn1,xn2,…,xnn,使an,xn1,xn2,…,xnn,an+1成等差數(shù)列,這樣可以得到新數(shù)列{bn}:a1,x11,a2,x21,x22,a3,x31,x32,x33,a4,…,an,設(shè)數(shù)列{bn}的前n項(xiàng)和為Sn,求S55(用數(shù)字作答,參考數(shù)據(jù):211=2048).解(1)當(dāng)n=1時(shí),a1=2;當(dāng)n≥2時(shí),an2n-1=a1+a22+a322+…+an2n-1-a1+所以an2n-1=2?an=2當(dāng)n=1時(shí),上式亦成立,所以an=2n.(2)由n+[1+2+3+…+(n-1)]=55?n=10.所以新數(shù)列{bn}前55項(xiàng)中包含數(shù)列{an}的前10項(xiàng),還包含x11,x21,x22,x31,x32,…,x98,x99,且x11=a1+a22,x21+x22=2(a2+a3)2,x31+x32+x33=3(a3所以S55