高考中的数列综合题选讲1.(2006陕西文、理)已知正项数列,其前n项和Sn满足10Sn=+5an+6,且a1,a3,a15成等比数列,求数列的通项an.解: 10Sn=an2+5an+6,①∴10a1=a12+5a1+6,解之得a1=2或a1=3.又10Sn-1=an-12+5an-1+6(n≥2),②由①-②得10an=(an2-an-12)+6(an-an-1),即(an+an-1)(an-an-1-5)=0 an+an-1>0,∴an-an-1=5(n≥2).当a1=3时,a3=13,a15=73.a1,a3,a15不成等比数列∴a1≠3;当a1=2时,a3=12,a15=72,有a32=a1a15,∴a1=2,∴an=5n-3.2.(2007山东理)设数列满足a1+3a2+32a3+…+3n-1an=.(Ⅰ)求数列的通项;(Ⅱ)设bn=,求数列的前n项和Sn.解:(I)验证时也满足上式,(II),,3.(2006全国Ⅰ卷理)设数列的前项的和,,(Ⅰ)求首项1a与通项na;(Ⅱ)设2nnnTS,,3,2,1n,证明:132niiT解:(Ⅰ)由Sn=an-×2n+1+,n=1,2,3,…,①得a1=S1=a1-×4+所以a1=2.再由①有Sn-1=an-1-×2n+,n=2,3,4,…将①和②相减得:an=Sn-Sn-1=(an-an-1)-×(2n+1-2n),n=2,3,…整理得:an+2n=4(an-1+2n-1),n=2,3,…,因而数列{an+2n}是首项为a1+2=4,公比为4的等比数列,即:an+2n=4×4n-1=4n,n=1,2,3,…,因而an=4n-2n,n=1,2,3,…,(Ⅱ)将an=4n-2n代入①得Sn=×(4n-2n)-×2n+1+=×(2n+1-1)(2n+1-2)=×(2n+1-1)(2n-1)Tn==×=×(-)所以,1niiT=1(ni-)=×(-)<4.(2005湖北文)设数列}{na的前n项和为Sn=2n2,}{nb为等比数列,且.)(,112211baabba(Ⅰ)求数列}{na和}{nb的通项公式;(Ⅱ)设nnnbac,求数列}{nc的前n项和Tn.解:(1):当;2,111San时,24)1(22,2221nnnSSannnn时当故{an}的通项公式为4,2}{,241daanann公差是即的等差数列.设{bn}的公比为.41,4,,11qdbqdbq则故.42}{,4121111nnnnnnbbqbb的通项公式为即(II),4)12(422411nnnnnnnbac]4)12(4)32(454341[4],4)12(45431[13212121nnnnnnnnTncccT两式相减得].54)56[(91]54)56[(314)12()4444(2131321nnnnnnnTnnT5.(1994全国文)设数列{an}的前n项和为Sn,若对于所有的自然数n,都有证明{an}是等差数列.证法一:令d=a2-a1.下面用数学归纳法证明an=a1+(n-1)d(n∈N).(1)当n=1时上述等式为恒等式a1=a1.当n=2时,a1+(2-1)d=a1+(a2-a1)=a2,等式成立.(2)假设当n=k(k≥2)时命题成立,ak=a1+(k-1)d.由题设,有Sk=21kaak,Sk+1=2111kaak,又Sk+1=Sk+ak+1∴(k+1)111122kkkaaakaa把ak=a1+(k-1)d代入上式,得(k+1)(a1+ak+1)=2ka1+k(k-1)d+2ak+1.整理得(k-1)ak+1=(k-1)a1+k(k-1)d. k≥2,∴ak+1=a1+kd.即当n=k+1时等式成立.由(1)和(2),等式对所有的自然数n成立,从而{an}是等差数列.证法二:当n≥2时,由题设,21111nnaanS,21nnaanS.所以an=Sn-Sn-1=21naan-2111naan同理有an+1=2111naan-21naan.从而an+1-an=2111naan-n(a1+an)+2111naan,整理得an+1-an=an-an-1=…=a2-a1从而{an}是等差数列.6.(2006福建文)已知数列na满足*12211,3,32().nnnaaaaanN(I)证明:数列1nnaa是等比数列;(II)求数列na的通项公式;(Ⅲ)若数列nb满足12111*44...4(1)(),nnbbbbnanN证明nb是等差数列。(I)证明:2132,nnnaaa21112*2112(),1,3,2().nnnnnnnnaaaaaaaanNaa1nnaa是以21aa2为首项,2为公比的等比数列。(II)解:由(I)得*12(),nnnaanN112211()()...()nnnnnaaaaaaaa12*22...2121().nnnnN(III)证明:1211144...4(1),nnbbbbna12(...)42,nnbbbnb122[(...)],nnbbbnnb①12112[(...)(1)](1).nnnbbbbnnb②②-①,得112(1)(1),nnnbnbnb...