求数列通项的常见类型和方法.doc

求数列通项的常见类型和方法一、公式法：(1)等差等比数列直接使用通项公式.2已知Sn=fn求an.用公式：an=S1n=1Sn-Sn-1n23已知Tn=a1a2an=fn.用公式：an=T1n=1TnTn-1n24已知Fan,Sn=0求an.有以下两种常见途径：途径一：Fan,Sn=0Sn=fanSn-1=fan-1an=fan-fan-1an(n2)途径二：Fan,Sn=0an=fSnSn-Sn-1=fSnSnan(n2)例1.已知：Sn=3n2-2n+1，求an.答：an=2n=16n-5n2例2.1已知：Sn=3+2an,求an.答：an=-32n-1.2已知：a1=1,an=2Sn22Sn-1n2,求an.答：an=1,n=1-24n2-8n+3,(n2)练：已知：a1=3,2an+1=SnSn+1,求an.答：an= 3 ,(n=1)183n-83n-5,(n2)二、作差作商法：1已知k1a1+k2a2+knan=fnfn可以换成an+1或Sn+1等,其中kn为一已知数列,求an.由得:k1a1+k2a2+kn-1an-1=fn-1,-:knan=fn-fn-1an=f1k1,(n=1)fn-fn-1kn,(n2)(2)已知k1a1+k2a2+knan=fnfn可以换成an+1或Sn+1等,kn为一已知数列,求an.由得:k1a1+k2a2+kn-1an-1=fn-1,-:knan=fn-fn-1an=k1f1,(n=1)knfn-fn-1,(n2)(3)已知(k1+a1)(k2+a2)(kn+an)=fnfn可以换成an+1等,其中kn为一已知数列,求an.由得:(k1+a1)(k2+a2)kn-1+an-1=fn-1,:an=f1-k1,(n=1)fnfn-1-kn,(n2)例3.04全国I理15已知：a1=1,a1+2a2+3a3+n-1an-1=ann2,求an.答:an=1,(n=1)n!2,(n2)练:07山东理17已知:a1+3a2+32a3+3n-1an=n3nN+,求an.答:an=13n.注：此类的特征为条件是关于an的“和的结构”或“积的结构”.若是“和的结构”则作差，若是“积的结构”则作商.三、累加累乘法：1已知an-an-1=fn,求an.用累加法得：an=a1+f2+f3+fn,(n2)2已知anan-1=fn,求an.用累乘法得：an=a1f2f3fn,(n2)3已知an+an-1=fn,求an.由an+an-1=fnan+1+an=fn+1,-得：an+1-an-1=fn+1-fn.从而a2n-1和a2n分别可用累加法求其通项.4已知anan-1=fn,求an.由anan-1=fnan+1an=fn+1,得：an+1an-1=fn+1fn.从而a2n-1和a2n分别可用累乘法求其通项.例4.1已知：a1=12,an+1=an+12n,求an.答：an=32-12n-1.2已知：a1=1,Sn=(n+1)an2,求an.答：an=n.四、分段类：已知：a1和an=fan-1,n为奇数gan-1,n为偶数,求an.由题有:a2n+1=ga2n=gfa2n-1a2n+2=fa2n+1=fga2n,从而数列a2n-1和a2n分别转化为一阶递推数列,再考虑用累加乘法、转化法、数归法等方法求解.推广1：推广到分三段的情形：已知：a1和an=fan-1,n=3kgan-1,n=3k+1han-1,n=3k+2,求an.由题有:a3n+3=fa3n+2=fha3n+1=fhga3na3n+2=ha3n+1=hga3n=hgfa3n-1a3n+1=ga3n=gfa3n-1=gfha3n-2 ,从而数列a3n-2,a3n-1和a3n分别化为一阶递推数列. 推广P：推广到分P段的情形：已知：a1和an=f1an-1,n=pk+1f2an-1,n=pk+2fman-1,n=pk+mfpan-1,n=pk,求an.记：Fix=fifi-1f2f1fpfp-1fi+2fi+1x,则有：apn+1=F1apn+1-papn+2=F2apn+2-papn+m=Fmapn+m-papn+p=Fpapn,数列apn+1,apn+2,apn+m,apn+p均分别化为一阶递推数列.例5.05北京例19改编设a1=a14,an+1=12an,n为偶数an+14,n为奇数,求an.解：a2n+1=12a2n=12a2n-1+14a2n+1-14=12a2n-1-14a2n-1-14为等比数列,首项为a1-14=a-14,公比为12a2n-1-14=a-1412n-1a2n-1=a-1412n-1+14同理:a2n+2=a2n+1+14=12a2n+14a2n+2-12=12a2n-12a2n-12为等比数列,首项为a2-12=a+14-12=a-14,公比为12a2n-12=a-1412n-1a2n=a-1412n-1+12综合,可得:an=a-1412n-12+14,n为奇数a-1412n2-1+12,n为偶数例6.已知：a1=67,an+1=2an,0an122an-1,12an0,d0,pN+求an.原式anp=an-1p+danp为等差数列,公差为d.4已知an=kan-1an-1+d,求an.原式1an=dk1an-1+1k,令bn=1an得:bn=dkbn-1+1k,化为五(1)推广：结论：设an=aan-1+bcan-1+dad-bc0,设p,q为方程x=ax+bcx+d的二根.1若p=q,则1an-p为等差数列;2若p q,则an-pan-q为等比数列.推广的注：线性递推数列an=aan-1+bcan-1+dad-bc0有时候为周期数列，其结论如下：1 若a=-d且a2+bc0，则T=2；2 若a=d且3a2+bc=0，则T=3；3 若a=d且a2+bc=0，则T=4；4 若a=d且5a4+10a2bc+b2c2=0，则T=5；5 若a=b=-c=12d0或c=d=-b=12a0，则T=6.5已知an=kan-1pan0,k0,pN+,求an.原式lgan=plgan-1+lgk,令bn=lgan,得:bn=pbn-1+lgk,化为五1.6已知an=kan-1+an+b,求an.法1：令an+cn+d=kan-1+cn-1+dan=kan-1+k-1cn+kd-kc-d,再令k-1c=a,kd-kc-db,得：c=ak-1,d=ka+b-bk-12,从而an+cn+d为等比数列.法2：原式ankn=an-1kn-1+an+bkn，再累加,但此时要用错位相减法，很麻烦.7已知an=kan-1+bn,求an.法1：原式ankn=an-1kn-1+bkn,令cn=ankncn=cn-1+bkn,再用累加法.法2：令an+pbn=kan-1+pbn-1an=kan-1+k-1pbn，令p=1k-1an+pbn为等比数列.例7.1已知：a1=1,an=12an-1+1,求an.答：an=2-12n-1.2已知：a1=1,an+1=3an+2n+1,求an.答：an=3n-n-1.3已知：a1=1,an+1=3an+2n,求an.答：an=3n-2n.4已知：在数列an中a1=1,an+1=1+1nan+n+12n,求an.09全国I理201改编解：an+1=1+1nan+n+12nan+1n+1=ann+12n,设bn=annbn+1=bn+12n.用累加法可得:bn=b1+12+12n-1=2-12n-1n2an=2n-n2n-1n2.经检验n=1时有a1=1=21-121-1，上式成立.an=2n-n2n-1nN+.六、数归法：例8.已知：a1=1,an0,n+1an+12-nan2+an+1an=0nN+,求an.(?)解：法1:令n=1解方程得：a2=12或-1舍;令n=2解方程得：a3=13或-12舍.从而猜测：an=1n.下面用数归法进行证明：1当n=1时:a1=1=11,公式成立.2假设当n=k时:有公式ak=1k成立,则当n=k+1时:将ak=1k带入n+1an+12-nan2+an+1an=0并化简得：kk+1ak+12+ak+1-1=0ak+1=1k+1或ak+1=-1k舍,公式成立.综上所述：对于一切正整数n均有:an=1n成立.法2：由n+1an+12-nan2+an+1an=0n+1(an+1an)2+an+1an-1=0an+1an=nn+1或an+1an=-1舍,an0anan-1=n-1n,再用累乘法可求得:an=1n.练：已知：a1=2,an+1=an2-nan+1,求an.02全国理221答：an=n+1例9.已知正数数列an的前n项和为Sn且Sn=12an+1an,求an.答:an=n-n-1,(nN+)解：法1：2Sn=an+1an2Sn=Sn-Sn-1+1Sn-Sn-1Sn+Sn-1=1Sn-Sn-1Sn2-Sn-12=1Sn2为等差数列,首项为S12=a12=1,公差为1.Sn2=nSn=nan=S1 n=1Sn-Sn-1n2=1, n=1n-n-1,n2=n-n-1,nN+.法2：由a1=1,a2=2-1,a3=3-2,故猜测:an=n-n-1.1当n=1时:a1=1-0=1,公式成立.2假设当n=k时:有ak=k-k-1成立,则当n=k+1时:Sk+1=12ak+1+1ak+1,而ak+1=Sk+1-Sk=12ak+1+1ak+1-12ak+1ak(此步是此题的难点，突破此难点的关键是同时使用条件中的Sn=12an+1an和隐含的an=Sn-Sn-1n2，从而找到ak+1与ak的关系.)ak+1-1ak+1=-ak+1ak=-k-k-1+1k-k-1=-2kak+12+2kak+1-1=0ak+1=k+1-k,公式成立.综上所述：对一切正整数n均有an=n-n-1成立.练：06全国II理22设数列an的前项和为Sn且方程x2-anx-an=0有一个根为Sn-1. 1)求a1,a2；2）求an.解：1)由题得：Sn-12-anSn-1-an=0令n=1得:S1-12-a1S1-1-a1=0a1-12-a1a1-1-a1=0a1=12分别令n=2,3,4可得：a2=16,a3=112,a4=120S1=12,S2=23,S3=34,S4=45从而猜测：Sn=nn+1,下面用数归法证明：(只证第二步)(此步是此题的难点：为什么不猜测an=1nn+1,而去猜测Sn=nn+1？原因是：若猜测an,则数归法的第二步中无法消去Sn,或者说ak+1与ak的关系不能找到.)2假设当n=k时有：Sk=kk+1则当n=k+1时；Sn-12-anSn-1-an=0Sn-12-Sn-Sn-1Sn-1-Sn-Sn-1=0Sn=12-Sn-1Sk+1=12-Sk=12-kk+1=k+1k+2等式也成立.下略.例10.07江西理22设正整数数列an满足：a2=4,对任意nN*有：2+1an+11an+1an+11n-1n+12+1an.1求a1,a3;2求数列an的通项an.解：(1)在原不等式中令n=1得:23a187a1=1;令n=2得:8a310a3=92于是猜测a