形如an2pan1qan型的递推式:用待定系数法,化为特殊数列{anan1}的形式求解。方法为:设an2kan1h(an1kan),比较系数得hkp,hkq,可解得h、k,于是{an1kan}是公比为h的等比数列,这样就化归为an1panq型。设递推公式为an1panqan1,其特征方程为x2pxq,即x2pxq01、若方程有两相异实根x1x2,则anAx1nBx2n;2、若方程有两相等实根x1x2,则a(ABn)xn.n1其中A,B可由初始条件确定。例2.1已知数列{an}满足a12,a23,an23an12an(nN*),求数列{an}的通项an方法一:令an2+an+1(an+1;于是=3,解得1=1或2=2an)=21=22=1an2an+12(an+1;an+1an是以为初项,2为公比的等比数列.an)1an+1an2n-1①an22an+1an+12an;an+12an是以1为初项,1为公比的等比数列(常数列).an+12an1②①②得an1+2n-1注:①式错位相消亦可解出an.方法二:作特征方程x23x+2,解得,=20x1=1x2anAx1nBx2nAB2n2A2BA1a12,a21,于是an=1+2n1.3,A4BB32例2.2已知数列{an}满足a1,a22,4a4aa(nN*),求数列{an}的通项an1n2n1n方法一:化简an2an11an,令an2+an+1(an+1an),4=11=2=12于是1,解得1=1=42=2an21an+11(an+11an);an21an+1是以3为初项,1为公比的等比数列222222an21an+1=3n1an22nan+1,nan+1是以1为初项,3为公差的等差数列.2n1,23222nan+113n,an+113nan3n22n,2n1.方法二:作特征方程4x24x10,解得x1x212n11an(ABn)x1=(ABn)()n11ABA=23n21(A2B),解得,于是an=n1.2B=322分式数列an1aanb通项公式cand分式线性递推数列an1aanb(a,b,c,dR,c0),其特征方程为xaxb,candcxd即cx2(da)xb0,1、若方程有两相异实根x1x2,则anx1成等比数列,其公比为acx1;anx2acx22、若方程有两相等实根x0,则1成等差数列,其公差为c.anx0acx0例.(1)设数列an满足a13,an17an2,求an.an4解:令x7x2,解得,=2x+4x1=1x2an7an216(an1)11=4an4anan127an225(an2)an4an4两式相除得an11=6an1an125an2a1是以a1=26n212为首项,为公比的等比数列.ana15an126)n1,an46n15n1an2(26n15n15(2)已知数列an满足a12,an21,nN*,求通项an.an1解:an21=2an112x1an,令xx,解得x011an1an1an11,取倒数得1=an1=11an1an1an11an11即11,1是以1为首项,为公差的等差数列.an=1an1=111an11a111n,an1,nN*an1nn