專題突破練6求數(shù)列的通項公式必備知識夯實練1.(2025江西贛州一模)已知數(shù)列{an}的前n項和為Sn,滿足3an=2Sn+1,則S5=()A.11 B.31 C.61 D.1212.(2025天津,6)已知數(shù)列{an}的前n項和為Sn,且Sn=-n2+8n,則{|an|}的前12項和為()A.112 B.48 C.80 D.643.(2025福建福州模擬)已知數(shù)列{an}滿足an+1=2an3an-1,且a1=2,A.512511 B.513C.256257 D.4.(2025山東臨沂一模)設(shè)數(shù)列{an}的前n項和為Sn,且Sn+nan=1,則滿足Sn>0.99時,n的最小值為()A.49 B.50 C.99 D.1005.(多選題)(2025山東濟(jì)南模擬)已知數(shù)列{an}的前n項和為Sn,且滿足a1=1,a2=3,an+1=3an-2an-1(n≥2),則下列說法正確的有()A.數(shù)列{an+1-an}為等差數(shù)列B.數(shù)列{an+1-2an}為等比數(shù)列C.an=2n-1D.Sn=2n+1-n-26.(2025河北張家口一模)已知數(shù)列{an}滿足a1=2,an>0且an+12-an27.(2025廣東廣州模擬)某農(nóng)村合作社引進(jìn)先進(jìn)技術(shù)提升某農(nóng)產(chǎn)品的深加工技術(shù),以此達(dá)到10年內(nèi)每年此農(nóng)產(chǎn)品的銷售額(單位:萬元)等于上一年的1.3倍再減去3.已知第一年(2024年)該公司該產(chǎn)品的銷售額為100萬元,則按照計劃該公司從2024年到2033年該農(nóng)產(chǎn)品的銷售總額約為萬元.(參考數(shù)據(jù):1.39≈10.6,1.310≈13.8,1.311≈17.9)8.(13分)若數(shù)列{an}的前n項和為Sn,且滿足Sn(Sn-an)+2an=0(n≥2),a1=2.(1)求證:{1Sn}(2)求數(shù)列{an}的通項公式.關(guān)鍵能力提升練9.(2025浙江湖州模擬)已知數(shù)列{an}滿足a1+3a2+9a3+…+3n-1an=n+13,設(shè)數(shù)列{an}的前n項和為Sn,若Sn<k恒成立,則實數(shù)k的最小值為(A.23 B.1C.76 D.10.(2025江蘇南通模擬)如圖,在楊輝三角形中,斜線l的上方,從1開始箭頭所示的數(shù)組成一個鋸齒形數(shù)列:1,3,3,4,6,5,10,…,記其前n項和為Sn,則S21=()A.351 B.360 C.361 D.35811.(2025江蘇宿遷模擬)設(shè)數(shù)列{an}的前n項和為Sn,且Sn=2an-2n,則()A.9a7>8a8 B.9a7<8a8C.9S7>7a8 D.9S7<7a812.(多選題)(2025山東德州模擬)對于數(shù)列{an},定義An=a1+2a2+…+2n-1ann為數(shù)列{an}的“好數(shù)”,已知某數(shù)列{an}的“好數(shù)”An=2n+1,記數(shù)列{an-kn}的前n項和為Sn,若Sn≤SA.2 B.167C.9742 D.13.(2025安徽銅陵模擬)已知數(shù)列{an}滿足2anan+1+an+1=3an,且a2=911,則使不等式1a1+1a2+…+1a核心素養(yǎng)創(chuàng)新練14.已知數(shù)列{an}滿足a1=1,an+1=an1+an(n∈N*).記數(shù)列{an}的前n項和為Sn,A.32<S100<3B.3<S100<4C.4<S100<92D.92<S100<

答案：1.D解析令n=1,得3a1=2S1+1=2a1+1,得a1=1.由3an=2Sn+1,當(dāng)n≥2時,3an-1=2Sn-1+1,兩式相減得3an-3an-1=2(Sn-Sn-1)=2an,即an=3an-1,即anan-1=3(n≥2),所以數(shù)列{an}是以a1=1為首項,3為公比的等比數(shù)列,所以S52.C解析由題意知a1=S1=7,當(dāng)n≥2時,an=Sn-Sn-1=-n2+8n+(n-1)2-8(n-1)=-2n+9.又a1=7適合上式,所以an=-2n+9.當(dāng)an>0時,n≤4,當(dāng)an<0時,n≥5.∴{|an|}前12項的和T12=S4-(S12-S4)=2S4-S12=2×(-16+32)-(-122+8×12)=80.故選C.3.A解析易知an≠0,從而由題意得1an+1=32-12·1an,即1an+1-1=-12(1an-1),1a1-1=-12≠0,所以數(shù)列1an-4.D解析因為Sn+nan=1,所以a1=12,當(dāng)n≥2時,Sn+nan=Sn-1+(n-1)an-1=1,所以(n+1)an=(n-1)an-1,即anan-1=n-1n+1(n≥2),此時an=anan-1故an=1n(n+1),Sn=1-nan=1-1n+1,若Sn=1-1n+1>0.99,解得n>99,又n5.BCD解析因為an+1=3an-2an-1(n≥2),所以an+1-an=2(an-an-1),又a2-a1=2≠0,則{an+1-an}是首項為2,公比為2的等比數(shù)列,故A錯誤;根據(jù)題意得an+1=3an-2an-1?an+1-2an=an-2an-1,又a2-2a1=1≠0,所以數(shù)列{an+1-2an}是首項為1,公比為1的等比數(shù)列,故B正確;由上得an+1-an=2n,所以an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+2+4+…+2n-1=1×(1-2n)1-Sn=(2+22+23+…+2n)-n=2×(1-2n)1-26.4-12n-1解析由題得,當(dāng)n≥2時,an2=(an2-an-12)+(an-12-an-22)+…+(a22-a12)+a12=12n-1+12n-2+…+12+(n-1)+7.3940解析該公司從2024年起的每年銷售額可構(gòu)成數(shù)列{an},n∈N*,n≤10,a1=100,依題意,當(dāng)n∈N*,n≤9時,an+1=1.3an-3,即an+1-10=1.3(an-10),a1-10=90≠0.因此數(shù)列{an-10}是首項為90,公比為1.3的等比數(shù)列,所以an-10=90×1.3n-1,即an=90×1.3n-1+10,則a1+a2+…+a10=90×(1-1.310)1-1.3+10×10所以從2024年到2033年該農(nóng)產(chǎn)品的銷售總額約為3940萬元.8.(1)證明當(dāng)n≥2時,an=Sn-Sn-1,且Sn(Sn-an)+2an=0.∴Sn[Sn-(Sn-Sn-1)]+2(Sn-Sn-1)=0,即SnSn-1+2(Sn-Sn-1)=0,即1Sn-故數(shù)列{1Sn}是首項為12,公差為(2)解由(1)知1S∴Sn=2n,當(dāng)n≥2時,an=Sn-Sn-1=-2當(dāng)n=1時,a1=2不適合上式.故an=29.D解析因為a1+3a2+9a3+…+3n-1an=n+1所以當(dāng)n≥2時,a1+3a2+9a3+…+3n-2an-1=n-兩式相減得3n-1an=13,所以an=當(dāng)n=1時,S1=a1=2當(dāng)n≥2時,Sn=23+19+127+…+1所以Sn=56-12(13)n,n≥2.當(dāng)n=1時,S1=23也符合上式,所以Sn=5由Sn<k恒成立,可得k≥56,所以k10.C解析當(dāng)n=2m-1(m∈N*)時,an=a2m-1=1+2+…+m=m(當(dāng)n=2m(m∈N*)時,an=a2m=m+2.綜上,S21=(=12[(12+22+…+112)+(1+2+…+11)]=12×11×12×236+111.B解析因為數(shù)列{an}的前n項和為Sn,且Sn=2an-2n,當(dāng)n=1時,則a1=S1=2a1-2,解得a1=2,當(dāng)n≥2且n∈N*時,由Sn=2an-2n可得Sn-1=2an-1-2n-1,上述兩式作差可得an=2an-2an-1-2n-1,整理可得an-2an-1=2n-1,等式an-2an-1=2n-1兩邊同時除以2n-1可得an2所以數(shù)列{an2n-1}是以a120所以an2n-1=2+n-1=n+1,所以an=(n+1)9a7=9×8×26=9×29,8a8=8×9×27=9×210,則9a7<8a8;Sn=2an-2n=(n+1)·2n-2n=n·2n,9S7=9×7×27=63×27,7a8=7×9×27=63×27,所以9S7=7a8.12.BCD解析因為An=a1+2a2+…+2n-1ann=2n+1,所以a1+2a2+…+2n-1當(dāng)n≥2時,a1+2a2+…+2n-2an-1=(n-1)·2n,兩式相減得2n-1an=n·2n+1-(n-1)·2n=(n+1)2n,所以an=2(n+1)(n≥2),當(dāng)n=1時,a1=4也符合上式.故an=2(n+1),則an-kn=(2-k)n+2,所以數(shù)列{an-kn}為等差數(shù)列.故Sn≤S6對任意的n(n∈N*)恒成立可化為a6-6k≥0,a7-7k≤0,即6(2-k)+2≥0,7(13.99解析由2anan+1+an+1=3an,a2=911可得2a1a2+a2=3a1?a1=3易知an≠0,兩側(cè)同時除以anan+1,可得2+1an=3an+1,整理得1an-1=3(1an+1-1),所以{1an則1an-1=23(13)n-1=故1a1-1+1a2-1+…+1an-1=23[1-(13)

n]1故1a1+1a2+…+1an易知f(n)=n+1-(13)n(n∈N*)單調(diào)遞增,則f(99)=100-1399<100<f(100)=101-1