【志鴻優(yōu)化設(shè)計(jì)】高考數(shù)學(xué)二輪專題升級訓(xùn)練 解答題專項(xiàng)訓(xùn)練(數(shù)列) 文（含解析） 新人教A版(1).doc

專題升級訓(xùn)練 解答題專項(xiàng)訓(xùn)練(數(shù)列)1.設(shè)數(shù)列an的前n項(xiàng)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(xiàng)公式.2.已知各項(xiàng)都不相等的等差數(shù)列an的前6項(xiàng)和為60,且a6為a1和a21的等比中項(xiàng).(1)求數(shù)列an的通項(xiàng)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(xiàng)和tn.3.已知數(shù)列an是公差為正的等差數(shù)列,其前n項(xiàng)和為sn,點(diǎn)(n,sn)在拋物線y=x2+x上;各項(xiàng)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(xiàng)公式;(2)記cn=anbn,求數(shù)列cn的前n項(xiàng)和tn.4.已知sn是等比數(shù)列an的前n項(xiàng)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(xiàng)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(xiàng)公式;(3)在(2)的條件下,求數(shù)列an前100項(xiàng)中的所有偶數(shù)項(xiàng)的和s.6.已知數(shù)列an(nn*)是首項(xiàng)為a,公比為q0的等比數(shù)列,sn是數(shù)列an的前n項(xiàng)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(xiàng)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(xiàng)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.(2013廣東深圳模擬,18)各項(xiàng)為正數(shù)的數(shù)列an滿足=4sn-2an-1(nn*),其中sn為an的前n項(xiàng)和.(1)求a1,a2的值;(2)求數(shù)列an的通項(xiàng)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,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(xiàng)為2,公差為3的等差數(shù)列.an=3n-1.又各項(xiàng)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.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(xiàng)的積為tn,tn=13q3(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=250-=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+2a+3a+(n-1)a+na,則-tn=a+2a+3a+(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,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q0,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=1024=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),即(an+1-an)(an+1+an)=2(an+1+an).數(shù)列an各項(xiàng)均為正數(shù),an+1+an0,an+1-an=2.數(shù)列an是首項(xiàng)為1,公差為2的等差數(shù)列,an=2n-1.(3)an=2n-1,a=