熱點(diǎn)2數(shù)列求和及等差、等比數(shù)列的綜合應(yīng)用年份202220232024角度題號(hào)角度題號(hào)角度題號(hào)新高考Ⅰ卷由遞推關(guān)系求通項(xiàng)公式與裂項(xiàng)相消法求和17求數(shù)列的通項(xiàng)公式與公差20數(shù)列的新定義問(wèn)題19新高考Ⅱ卷等差、等比數(shù)列的綜合應(yīng)用17數(shù)列的通項(xiàng)公式與前n項(xiàng)和的綜合應(yīng)用18數(shù)列的應(yīng)用19【典例1】(13分)(規(guī)范解答)(2024·全國(guó)甲卷)已知數(shù)列{an}的前n項(xiàng)和為Sn,且4Sn=3an+4.(1)求{an}的通項(xiàng)公式;(2)設(shè)bn=(1)n1nan,求數(shù)列{bn}的前n項(xiàng)和Tn.【審題思維】(1)由已知遞推關(guān)系進(jìn)行轉(zhuǎn)化,然后結(jié)合等比數(shù)列的通項(xiàng)公式即可求解;(2)先求出bn,然后結(jié)合錯(cuò)位相減法求解.【解析】(1)因?yàn)?Sn=3an+4,所以4Sn+1=3an+1+4,兩式相減可得4an+1=3an+13an, ……3分即an+1=3an,又因?yàn)?S1=3a1+4,所以a1=4,故數(shù)列{an}是首項(xiàng)為4,公比為3的等比數(shù)列,所以an=4·(3)n1; …6分(2)bn=(1)n1nan=4n·3n1, ……8分所以Tn=4(1·30+2·31+3·32+…+n·3n1),3Tn=4(1·31+2·32+3·33+…+n·3n), ……10分兩式相減可得:2Tn=4(1+31+32+…+3n1n·3n)=4(1-3n-2n·3n)=(2所以Tn=(2n1)3n+1. ……13分【題后反思】本題考查由遞推關(guān)系求數(shù)列的通項(xiàng)公式、錯(cuò)位相減法的應(yīng)用,解題時(shí)謹(jǐn)防兩個(gè)易錯(cuò)點(diǎn):(1)由Sn求an,注意對(duì)n=1時(shí)對(duì)應(yīng)的首項(xiàng)的檢驗(yàn);(2)利用錯(cuò)位相減法求解時(shí),不要弄錯(cuò)項(xiàng)數(shù).【典例2】(2024·泉州二模)已知數(shù)列{an}和{bn}的各項(xiàng)均為正,且a3=18b1,{bn}是公比為3的等比數(shù)列.數(shù)列{an}的前n項(xiàng)和Sn滿足4Sn=an2+2a(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=bn+3(bn+3-3)(bn+3-1)+a【審題思維】(1)利用遞推公式可證得數(shù)列{an}是等差數(shù)列,可求出數(shù)列{an}的通項(xiàng)公式;利用等比數(shù)列的性質(zhì),可求出{bn}的通項(xiàng)公式;(2)求出cn,根據(jù)裂項(xiàng)相消法和分組求和法求Tn.【解析】(1)由題設(shè),當(dāng)n=1時(shí),4S1=a12+2a1,所以a1=2或a1=0(由4Sn=an2+2an,知4Sn1=an-12+2an1,兩式相減得(an+an1)(anan12)=0,所以an+an1=0(舍)或anan12=0,即所以數(shù)列{an}是首項(xiàng)為2,公差為2的等差數(shù)列,所以an=2n.又a3=18b1=6,所以b1=13,所以bn=3n2(2)cn=bn+3(bn+3-3)(bn+3-1)+ancos=12(13n-113n則Tn=12[(13-1132-1)+(132-113當(dāng)n為偶數(shù)時(shí),Tn=12(1213n當(dāng)n為奇數(shù)時(shí),Tn=12(1213n+1-1)所以Tn=12【題后反思】本題考查等差、等比數(shù)列通項(xiàng)公式的求法及裂項(xiàng)相消法、分組求和法,解題時(shí)需要注意:(1)利用裂項(xiàng)相消法求和時(shí),要保證相消前后等號(hào)成立;(2)當(dāng)通項(xiàng)公式中出現(xiàn)(1)n,(1)n+1,求和時(shí)常按奇偶討論.1.★★★☆☆(2024·成都模擬)記數(shù)列{an}的前n項(xiàng)和為Sn,已知2Sn=n2+an+a11.(1)若a1≠1,證明:{ann}是等比數(shù)列;(2)若a2是a1和a3的等差中項(xiàng),設(shè)bn=1anan+2,求數(shù)列{bn}的前【解析】(1)對(duì)2Sn=n2+an+a11①,當(dāng)n≥2時(shí),有2Sn1=(n1)2+an1+a11②,①②:2(SnSn1)=2n1+anan1,即2an=2n1+anan1,經(jīng)整理,可得ann=(1)[an1(n1)],a1≠1,故{ann}是以a11為首項(xiàng)、1為公比的等比數(shù)列.(2)由(1)知ann=(1)n1(a11),有a2=3a1,a3=a1+2,由題設(shè)知2a2=a1+a3,即2(3a1)=a1+(a1+2),則a1=1,故an=n.而bn=1anan+2=1n(Tn=b1+b2+…+bn1+bn=12(1113+1214+…+1n-11n+1+1n1故Tn=3412(1n+12.★★★☆☆(2024·紹興三模)已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,Sn=nn+2an+1,設(shè)bn=(1)求證:數(shù)列{bn}為等比數(shù)列;(2)求數(shù)列{Sn}的前n項(xiàng)和Tn.【解析】(1)Sn=nn+2an+1=nn+2(Sn+1Sn),即(n+2)Sn=n(Sn+1Sn),即nSn+1=(2n+2)Sn,則nSn+1n即bn+1=2bn,又b1=S11=a故數(shù)列{bn}是以2為首項(xiàng)、以2為公比的等比數(shù)列;(2)由(1)得bn=2n,即Snn=2n,則Sn=n·2則Tn=1·21+2·22+…+n·2n,有2Tn=1·22+2·23+…+n·2n+1,則Tn2Tn=Tn=2+22+23+…+2nn·2n+1=2(1-2n)1-2n·2n+1=2n+12n·2n+1=(1n)·2n+12,故3.★★★☆☆(2024·安康模擬)記Sn為數(shù)列{an}的前n項(xiàng)和,已知a1=1,nSn+1(n+1)Sn=n2+n.(1)求{an}的通項(xiàng)公式;(2)若bn=(1)nan+(-1)n+12n,求數(shù)列{bn}的前2n項(xiàng)和【解析】(1)由nSn+1(n+1)Sn=n2+n,可得nSn+1(n+1)Sn=n(n+1),所以Sn+1n+1Snn=1,又由a1=1,所以S11=a11=1,所以Snn=1+(n1)×1=n,則Sn=n當(dāng)n≥2時(shí),Sn1=(n1)2,所以SnSn1=an=n2(n1)2=2n1,又當(dāng)n=1時(shí),a1=1滿足上式,所以{an}的通項(xiàng)公式為an=2n1.(2)由(1)可知,當(dāng)n為奇數(shù)時(shí),bn=an=12n;當(dāng)n為偶數(shù)時(shí),bn=an+2×2n=2n1+2n+1,所以T2n=b1+b2+b3+b4+…+b2n=(b1+b3+b5+…+b2n1)+(b2+b4+b6+…+b2n)=2n+23+25+27+29+…+22n+1=2n+8(1+22+24+26+…+22n2)=2n+8×1-4n1-4=2n+4.★★★★☆(2024·上海模擬)已知f(x)=12x2+12x,數(shù)列{an}的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)(n∈N*)均在函數(shù)y=f(x)(1)求數(shù)列{an}的通項(xiàng)公式;(2)若g(x)=4x4x+2,令bn=g(an2025)(n∈N*),求數(shù)列{b【解析】(1)因?yàn)辄c(diǎn)(n,Sn)(n∈N*)均在函數(shù)f(x)=12x2+12x所以Sn=12n2+12當(dāng)n=1時(shí),S1=12+12=1,即a當(dāng)n≥2時(shí),an=SnSn1=12n2+12n[12(n1)2+12(n1)]=12n2+12n(12n2n+1因?yàn)閍1=1滿足上式,所以an=n;(2)因?yàn)間(x)=4x所以g(x)+g(1x)=4x4x+2+41-x41因?yàn)閍n=n,所以bn=g(an2025)=g(n2所以T2024=b1+b2+b3+…+b20