2021新高考數(shù)學二輪總復習專題突破練15求數(shù)列的通項及前n項和

專題突破練15求數(shù)列的通項及前n項和1.已知等差數(shù)列{an}的前n項和為Sn,且a1=1,S3+S4=S5.(1)求數(shù)列{an}的通項公式;(2)令bn=(1)n1an,求數(shù)列{bn}的前n項和Tn.2.(2020山東濱州二模,18)已知{an}為等差數(shù)列,a3+a6=25,a8=23,{bn}為等比數(shù)列,且a1=2b1,b2b5=a11.(1)求{an},{bn}的通項公式;(2)記cn=an·bn,求數(shù)列{cn}的前n項和Tn.3.(2020全國Ⅲ,理17)設數(shù)列{an}滿足a1=3,an+1=3an4n.(1)計算a2,a3,猜想{an}的通項公式并加以證明;(2)求數(shù)列{2nan}的前n項和Sn.4.(2020山東聊城二模,17)已知數(shù)列{an}的各項均為正數(shù),其前n項和為Sn,且an2+an=2Sn+34(n∈N(1)求數(shù)列{an}的通項公式;(2)若bn=1Sn,求{bn}的前n項和T5.(2020山東青島5月模擬,17)設數(shù)列{an}的前n項和為Sn,a1=1,,

給出下列三個條件;條件①:數(shù)列{an}為等比數(shù)列,數(shù)列{Sn+a1}也為等比數(shù)列;條件②:點(Sn,an+1)在直線y=x+1上;條件③:2na1+2n1a2+…+2an=nan+1.試在上面的三個條件中任選一個,補充在上面的橫線上,完成下列兩問的解答:(1)求{an}的通項公式;(2)設bn=1log2an+1·log2an6.(2020山東菏澤一模,18)已知數(shù)列{an}滿足nan+1(n+1)an=1(n∈N*),且a1=1.(1)求數(shù)列{an}的通項公式;(2)若數(shù)列{bn}滿足bn=an3n-1,求數(shù)列{bn}的前7.已知數(shù)列{an}的前n項和Sn=3n2+8n,{bn}是等差數(shù)列,且an=bn+bn+1.(1)求數(shù)列{bn}的通項公式;(2)令cn=(an+1)n+1(bn+2)n8.(2020天津南開區(qū)一模,18)已知數(shù)列{an}的前n項和Sn=n2+n2,數(shù)列{bn}滿足:b1=b2=2,bn+1bn=2n+1(n∈(1)求數(shù)列{an},{bn}的通項公式;(2)求∑i=1naib2i-1專題突破練15求數(shù)列的通項及前n項和1.解(1)設等差數(shù)列{an}的公差為d,由S3+S4=S5可得a1+a2+a3=a5,即3a2=a5,∴3(1+d)=1+4d,解得d=2.∴an=1+(n1)×2=2n1.(2)由(1)可得bn=(1)n1·(2n1).∴T2n=13+57+…+(2n3)(2n1)=(2)×n=2n.∴當n為偶數(shù)時,Tn=n;當n為奇數(shù)時,Tn=Tn1+bn=(n1)+(1)n1an=(n1)+(1)n1(2n1)=(n1)+(2n1)=n.綜上,Tn=(1)n+1n.2.解(1)設等差數(shù)列{an}的公差為d,由題意得2a1所以數(shù)列{an}的通項公式an=3n1.設等比數(shù)列{bn}的公比為q,由a1=2b1,b2b5=a11,得b1=1,b12q5=32,解得所以數(shù)列{bn}的通項公式bn=2n1.(2)由(1)知,cn=anbn=(3n1)×2n1,則Tn=c1+c2+c3+…+cn1+cn=2×20+5×21+8×22+…+(3n4)×2n2+(3n1)×2n1,2Tn=2×21+5×22+8×23+…+(3n4)×2n1+(3n1)×2n.兩式相減得Tn=2+3(21+22+…+2n1)(3n1)×2n=2+3×2-2n-1×21-2(3n1)×2n=4+(43n)×2n,所以T3.解(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1(2n+3)=3[an(2n+1)],an(2n+1)=3[an1(2n1)],……a25=3(a13).因為a1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以Sn=3×2+5×22+7×23+…+(2n+1)×2n.①從而2Sn=3×22+5×23+7×24+…+(2n+1)×2n+1.②①②得Sn=3×2+2×22+2×23+…+2×2n(2n+1)×2n+1.所以Sn=(2n1)2n+1+2.4.解(1)因為an2+an=2Sn+34(n∈N*所以當n≥2時,an-12+an1=2Sn1+①②得an2-an-12+anan1即an2-an-1所以(an+an1)(anan11)=0.因為an>0,所以anan1=1,所以數(shù)列{an}是公差為1的等差數(shù)列,當n=1時,由a12+a1=2S1+34可得,a1=32,所以an=a1+(n1)d=32+(n(2)由(1)知Sn=na1+n(n-1)2d=n(n+2)2,所以bn=1Sn=2n(n+2)=1n-1n5.解(1)方案一:選條件①.因為數(shù)列{Sn+a1}為等比數(shù)列,所以(S2+a1)2=(S1+a1)(S3+a1),即(2a1+a2)2=2a1(2a1+a2+a3).設等比數(shù)列{an}的公比為q,因為a1=1,所以(2+q)2=2(2+q+q2),解得q=2或q=0(舍).所以an=a1qn1=2n1(n∈N*).方案二:選條件②.因為點(Sn,an+1)在直線y=x+1上,所以an+1=Sn+1(n∈N*),所以an=Sn1+1(n≥2).兩式相減得an+1an=an,an+1an=2(n≥2).因為a1=1,a2=S1+1=a1+1=2,a所以數(shù)列{an}是首項為1,公比為2的等比數(shù)列,所以an=a1qn1=2n1(n∈N*).方案三:選條件③.當n≥2時,因為2na1+2n1a2+…+2an=nan+1(n∈N*),①所以2n1a1+2n2a2+…+2an1=(n1)an.所以2na1+2n1a2+…+22an1=2(n1)an.②①②得2an=nan+12(n1)an,即an+1an=2(當n=1時,2a1=a2,a2a1=所以數(shù)列{an}是首項為1,公比為2的等比數(shù)列,所以an=a1qn1=2n1(n∈N*).(2)由(1)得an=2n1(n∈N*),所以bn=1log2an+1·log2an+3=1n(n+2)=121n-6.解(1)因為nan+1(n+1)an=1,所以an所以ann-aan…a22-a所以anna1=11n(n又a1=1,所以an所以an=2n1(n≥2).又a1=1,也符合上式,所以對任意正整數(shù)n,an=2n1.(2)結(jié)合(1)得bn=2n-13n-1,所以Sn=113Sn=13+332+①②,得23Sn=1+213+132+…+13n-12n-7.解(1)由題意知當n≥2時,an=SnSn1=6n+5,當n=1時,a1=S1=11,所以an=6n+5.設數(shù)列{bn}的公差為d,由a1=b1+b2,a2=b2+b(2)由(1)知cn=(6n+6)n+1(3n+3)n=3(n+1)·2n+1,又Tn=c1+c2+c3+…+cn,得Tn=3×[2×22+3×23+4×24+…+(n+1)×2n+1],2Tn=3×[2×23+3×24+4×25+…+(n+1)×2n+2],兩式作差,得Tn=3×[2×22+23+24+…+2n+1(n+1)×2n+2]=3×4+4(2n-1)2-1(n+1)×2n+28.解(1)當n≥2時,an=SnSn1=n2+n=1時,a1=S1=1,滿足上式,∴an=n.∵bn+1bn=2n+1,∴bnbn1=2n(n≥2),∴bn+1=2bn1(n≥2),∴數(shù)列{bn}的奇數(shù)項和偶數(shù)項分別是2為首項,2為公比的等比數(shù)列,∴bn=2(2)∑i=1naib2i-