文档介绍：高考数列常考题型归纳总结类型1an+1=an+f(n解法:把原递推公式转化为an+1-an=f(n,利用累加法(逐差相加法求解。例:已知数列{an}满足a1=解:由条件知:an+1-an=12,an+1=an+1=11n+n2,求an。-1n+1n+n2n(n+1=1n分别令n=1,2,3,⋅⋅⋅⋅⋅⋅,(n-1,代入上式得(n-1个等式累加之,即(a2-a1+(a3-a2+(a4-a3+⋅⋅⋅⋅⋅⋅+(an-an-1=(1-12+(12-13+(1n13-14+⋅⋅⋅⋅⋅⋅+(1n-1-1n所以an-a1=1-a1=1212+1-1n=32-1n,∴an=类型2an+1=f(nan解法:把原递推公式转化为23an+1an=f(n,利用累乘法(逐商相乘法求解。nn+1例:已知数列{an}满足a1=解:由条件知之,即a2a1∙a3a2∙a4a323,an+1=an,求an。an+1an=nn+1,分别令n=1,2,3,⋅⋅⋅⋅⋅⋅,(n-1,代入上式得(n-1个等式累乘∙⋅⋅⋅⋅⋅⋅∙anan-123n=12⨯23⨯34⨯⋅⋅⋅⋅⋅⋅⨯n-1n⇒ana1=1n又a1=,∴an=例:已知a1=3,an+1=解:an=3(n-1-13(n-1+23n-43n-13n-13n+2an(n≥1,求an。∙3(n-2-13(n-2+27⋅4∙⋅⋅⋅∙3⨯2-13⨯2+26∙3-13+2a1=⋅3n-3n-52⋅⋅3=85n-3。1变式:(2004,全国I,理15.)已知数列{an},满足a1=1,an=a1+2a2+3a3+⋅⋅⋅+(n-1an-1(n≥2,则{an}的通项an=⎨⎧1⎩___n=1n≥2解:由已知,得an+1=a1+2a2+3a3+⋅⋅⋅+(n-1an-1+nan,用此式减去已知式,得当n≥2时,an+1-an=nan,即an+1=(n+1an,又a2=a1=1,∴a1=1,a2a1=1,a3a2=3,a4a3=4,⋅⋅⋅,anan-1=n,将以上n个式子相乘,得an=n!2(n≥2类型3an+1=pan+q(其中p,q均为常数,(pq(p-1≠0)。解法(待定系数法):把原递推公式转化为:an+1-t=p(an-t,其中t=换元法转化为等比数列求解。例:已知数列{an}中,a1=1,an+1=2an+3,:设递推公式an+1=2an+3可以转化为an+1-t=2(an-t即an+1=2an-t⇒t=-+1+3=2(an+3,令bn=an+3,则b1=a1+3=4,且bn+1bn=an+1+3an+3=-p,再利用n-1n+1=2所以{bn}是以b1=4为首项,2为公比的等比数列,则bn=4⨯2,所以an=2n+1-:(2006,重庆,文,14)在数列{an}中,若a1=1,an+1=2an+3(n≥1,则该数列的通项an=_______________(key:an=2n+1-3)变式:()已知数列{an}满足a1=1,an+1=2an+1(n∈N*.(I)求数列{an}的通项公式;(II)若数列{bn}滿足4b-14b12-14bn-1=(an+1n(n∈N,证明:数列{bn}是等差数列;b*(Ⅲ)证明:n12-13<aa+a2<n∈N*.2a+...+an3an+12(n(I)解:an+1=2an+1(n∈N*,∴an+1+1=2(an+1,∴{an+1}是以a1+1=2为首项,2为公比的等比数列∴ann+1=*n=2-1(n∈N.(II)证法一:4k1-14k2-1...4kn-1=(an+1kn.∴4(k1+k2+...+kn-n=2nkn.∴2[(b1+b2+...+bn-n]=nbn,2[(b1+b2+...+bn+bn+1-(n+1]=(n+1bn+1.②-①,得2(bn+1-1=(n+1bn+1-nbn,即(n-1bn+1-nbn+2=0,nbn+2-(n+1bn+1+2=0.③-④,得nbn+2-2nbn+1+nbn=0,即bn+2-2bn+1+bn=0,∴bn+2-bn+1=bn+1-bn(n∈N*,∴{bn}是等差数列①②证法二:同证法一,得(n-1bn+1-nbn+2=0令n=1,得b1==2+d(d∈R,下面用数学归纳法证明bn=2+(n-1d.(1)当n=1,2时,等式成立