高考數(shù)學(xué)精品復(fù)習(xí)資料 2019.5專題升級訓(xùn)練 解答題專項訓(xùn)練(數(shù)列)1.設(shè)數(shù)列an的前n項和Sn滿足2Sn=an+1-2n+1+1,nN*,且a1,a2+5,a3成等差數(shù)列.(1)求a1的值;(2)求數(shù)列an的通項公式.2.已知各項都不相等的等差數(shù)列an的前6項和為60,且a6為a1和a21的等比中項.(1)求數(shù)列an的通項公式;(2)若數(shù)列bn滿足bn+1-bn=an(nN*),且b1=3,求數(shù)列的前n項和Tn.3.已知數(shù)列an是公差為正的等差數(shù)列,其前n項和為Sn,點(n,Sn)在拋物線y=x2+x上;各項都為正數(shù)的等比數(shù)列bn滿足b1b3=,b5=.(1)求數(shù)列an,bn的通項公式;(2)記Cn=anbn,求數(shù)列Cn的前n項和Tn.4.已知Sn是等比數(shù)列an的前n項和,S4,S10,S7成等差數(shù)列.(1)求證a3,a9,a6成等差數(shù)列;(2)若a1=1,求數(shù)列的前n項的積.5.已知數(shù)列an滿足:a1=1,an+1=(1)求a2,a3;(2)設(shè)bn=a2n-2,nN*,求證:bn是等比數(shù)列,并求其通項公式;(3)在(2)的條件下,求數(shù)列an前100項中的所有偶數(shù)項的和S.6.已知數(shù)列an(nN*)是首項為a,公比為q0的等比數(shù)列,Sn是數(shù)列an的前n項和,已知12S3,S6,S12-S6成等比數(shù)列.(1)當(dāng)公比q取何值時,使得a1,2a7,3a4成等差數(shù)列?(2)在(1)的條件下,求Tn=a1+2a4+3a7+na3n-2.7.已知數(shù)列an的各項排成如圖所示的三角形數(shù)陣,數(shù)陣中每一行的第一個數(shù)a1,a2,a4,a7,構(gòu)成等差數(shù)列bn,Sn是bn的前n項和,且b1=a1=1,S5=15. (1)若數(shù)陣中從第三行開始每行中的數(shù)按從左到右的順序均構(gòu)成公比為正數(shù)的等比數(shù)列,且公比相等,已知a9=16,求a50的值;(2)設(shè)Tn=+,求Tn.8.(20xx·廣東深圳模擬,18)各項為正數(shù)的數(shù)列an滿足=4Sn-2an-1(nN*),其中Sn為an的前n項和.(1)求a1,a2的值;(2)求數(shù)列an的通項公式;(3)是否存在正整數(shù)m,n,使得向量a=(2an+2,m)與向量b=(-an+5,3+an)垂直?說明理由.#1.解:(1)在2Sn=an+1-2n+1+1中,令n=1,得2S1=a2-22+1,令n=2,得2S2=a3-23+1,來源:解得a2=2a1+3,a3=6a1+13.又2(a2+5)=a1+a3,解得a1=1.(2)2Sn=an+1-2n+1+1,2Sn+1=an+2-2n+2+1,得an+2=3an+1+2n+1,又a1=1,a2=5也滿足a2=3a1+21,an+1=3an+2n對nN*成立.an+1+2n+1=3(an+2n),an+2n=3n,an=3n-2n.2.解:(1)設(shè)等差數(shù)列an的公差為d(d0),則解得an=2n+3.(2)由bn+1-bn=an,bn-bn-1=an-1(n2,nN*),bn=(bn-bn-1)+(bn-1-bn-2)+(b2-b1)+b1=an-1+an-2+a1+b1=(n-1)·+3=n(n+2).bn=n(n+2)(nN*).,Tn=.3.解:(1)Sn=n2+n,當(dāng)n=1時,a1=S1=2;當(dāng)n2時,Sn-1=(n-1)2+(n-1)=n2-n+1.an=Sn-Sn-1=3n-1(n2).當(dāng)n=1時,a1=3-1=2滿足題意.數(shù)列an是首項為2,公差為3的等差數(shù)列.an=3n-1.又各項都為正數(shù)的等比數(shù)列bn滿足b1b3=,b5=,b2=b1q=,b1q4=,解得b1=,q=,bn=.(2)Cn=(3n-1)×,Tn=2×+5×+(3n-4)×+(3n-1)×,Tn=2×+5×+(3n-4)×+(3n-1)×,-,得Tn=1+3-(3n-1)×=1+3×-(3n-1)×-3×-(3n-1)×.Tn=5-.4.解:(1)當(dāng)q=1時,2S10S4+S7,q1.由2S10=S4+S7,得.a10,q1,2q10=q4+q7.來源:則2a1q8=a1q2+a1q5.2a9=a3+a6.a3,a9,a6成等差數(shù)列.(2)依題意設(shè)數(shù)列的前n項的積為Tn,Tn=··=13·q3·(q2)3··(qn-1)3=q3·(q3)2··(q3)n-1=(q3)1+2+3+(n-1)=(q3.又由(1)得2q10=q4+q7,2q6-q3-1=0,解得q3=1(舍),q3=-.Tn=.5.解:(1)a2=,a3=-.(2)=,又b1=a2-2=-,數(shù)列bn是等比數(shù)列,且bn=-.(3)由(2)得a2n=bn+2=2-(n=1,2,3,50),S=a2+a4+a100=2×50-=100-1+=99+.6.解:(1)由題意可知,a0.當(dāng)q=1時,則12S3=36a,S6=6a,S12-S6=6a,此時不滿足條件12S3,S6,S12-S6成等比數(shù)列;當(dāng)q1時,則12S3=12×,S6=,S12-S6=,由題意得12×,化簡整理得(4q3+1)(3q3-1)(1-q3)(1-q6)=0,解得q3=-,或q3=,或q=-1.當(dāng)q=-1時,a1+3a4=-2a,2a7=2a,a1+3a42(2a7),不滿足條件;當(dāng)q3=-時,a1+3a4=a(1+3q3)=,2(2a7)=4aq6=,即a1+3a4=2(2a7),當(dāng)q=-時,滿足條件;當(dāng)q3=時,a1+3a4=a(1+3q3)=2a,2(2a7)=4aq6=,a1+3a42(2a7),從而當(dāng)q3=時,不滿足條件.綜上,當(dāng)q=-時,使得a1,2a7,3a4成等差數(shù)列.(2)由(1)得na3n-2=na.Tn=a+2×a+3×a+(n-1)·a+na,則-Tn=a+2×a+3×a+(n-1)a+na,-得Tn=a+a+a+a+a-na=a-a,Tn=a-a.7.解:(1)bn為等差數(shù)列,設(shè)公差為d,b1=1,S5=15,來源:S5=5+10d=15,d=1.bn=1+(n-1)×1=n.設(shè)從第3行起,每行的公比都是q,且q>0,a9=b4q2,4q2=16,q=2,1+2+3+9=45,故a50是數(shù)陣中第10行第5個數(shù),而a50=b10q4=10×24=160.(2)Sn=1+2+n=,Tn=+=+=2=2.8.解:(1)當(dāng)n=1時,=4S1-2a1-1,即(a1-1)2=0,解得a1=1,當(dāng)n=2時,=4S2-2a2-1=4a1+2a2-1=3+2a2,解得a2=3或a2=-1(舍去).(2)由已知=4Sn-2an-1,=4Sn+1-2an+1-1,-得=4an+1-2an+1+2an=2(an+1+an),來源:即(an+1-an)(an+1+an)=2(an+1+an).數(shù)列an各項均為正數(shù),an+1+an>0,an+1-an=2.數(shù)列an是首項為1,公差為2的等差數(shù)列,an=2n-1.(3)an=2n-1,a=(2an+2,m)=(2(2n+3),m)0,b=(-an+5,3+an)=(-(2n+9),2(n+1)0.又aba·b=0m(n+1)-(2n+3)(2n+9)=0m=4(n+1)+16+,m,nN*,n+1=7,m=4×7+16+1,即n=6,m=45.來源:當(dāng)且僅當(dāng)n=6,m=45時,ab.