高中理科数学解题方法篇_数列_

1高中数列解题方法及综合学校慈济中学姓名晋春2高考递推数列分类类型1渗角函数周期性数列角函数的结合是一类创新试题利用角函数的周期性体数列的变利用角等式进行放缩是证明数列等式的常见方法醔例12008湖南卷18满12数列{an}满足a1=1a2=2222(1cos)sin,1,2,3...22nnnnaanππ+=++=求a3a4,并求数列{an}的通式2212311224222221212121212121:1,2,(1cos)sin12,22(1cos)sin24(21)2121(),[1cos]sin221,1{}112()kkkkkkkaaaaaaaakknkkNaaaaaaaknkkNππππππ∗+−−+−−−∗===++=+==++==−−=−∈=++=+−===∈


解因为所一般地时即所数列是首为差为的等差数列因时22222222222,[1cos]sin222{}2221,21()2{}2,2()kkkkkknnnkkaaaaannkkNaankkNππ+∗∗=++==+=−∈==∈

所数列是首为比为的等比数列因故数列的通式为本题为两种情况采取非常规的递推数列求通的方法利用角函数的诱式找递推关系体角函数的周期性进而求出该数列的通为一段数列醔例22009江西文21满12数列{an}的通)3sin3(cos222ππnnnan−=其前n和为1求sn2nnnnsb43⋅=求数列{bn}的前n和Tn361323212)13(2)13()94(2)94(2)49(2518...231213])3(2)13()23([...)6254()3221()(...)()(,32cos3sin3cos)1(:221313233313222222222231323654321322−−−=−=−+−+−=−=−=−=+=−+++=+−+−−++++−+++−=+++++++++==−−−−−−−kkkkkkasskkasskkkkkkaaaaaaaaasnnnkkkkkkkkkk故由于解πππ12321232123232313829218)449411494913(21)44949...4913(213)1449...42213(214)449...422413(2124494)2()(,3,6)43(13,6)31)(1(23,613+−+−−∗−⋅−=−−=+−−−+=+−+++=−++++=++++=+=⋅=∈=+−=−+−=−−=nnnnnnnnnnnnnnnnnnnnTnnnTnTnTnnsbNkknnnknnnknns故两式相得故例32009江西理85数列{an}的通)3sin3(cos222ππnnnan−=其前n和为sn则sn为A470B490C495D510类型2an+1=an+f(n)解法思路把原递推式转为an+1-an=f(n)利用累加法差相加法求解例42008江西理5在数列{an}中a1=2,an+1=an+ln)11(n+,则an=4A2+lnnB2+(n-1)lnnC2+nlnnD1+n+lnn例52009全I理22在数列{an}中a1=1,an+1=nnnan21)11(+++1设nabnn=求数列{an}的通式2求数列{an}的前n和醔...21,21212111,1)1(:2312111+=+=+=+=++==+bbbbbbnanaabnnnnnn而即且由已知得解422)1()1()2(224222,22)212()1()2(2121)2(21221...2121)2(2111110212121111112111121−+++=+=+−=−=−===−=−=−==≥−=++++=≥+=−=−−==−−−−−−−∑∑∑∑−−−nnnknnnkknnnknnkknnnnnnnnnnnnnnnsnnknnTTTTTnnnabbnbbnbbkkk所又于是则知由故所求通式为又于是类型3an+1=f(n)an解法思路把原递推式转为)(1nfaann=+利用累乘法商相乘法求解例62004全I理15已知数列{an}满足a1=1,an=a1+2a2+3a3+…+(n1)an1(n鄁2)则{an}的通an=_____解由已知得an+1=a1+2a2+3a3+…+(n1)an1+nan用式去已知式得n鄁2时an+1an=nan即an+1=n+1an又a2=a15)2(2!,,...4,3,,1,113423121≥======−nna，nnaaaaaaaaannn得个式子相乘将所类型4an+1=pan+q其中p醓q均为常数且pq(p1)≠0解法思路待定系数法把原递推式转为an+1t=p(ant)其中pqt−=1再利用换元法转为等比数列求解或转为队循数列来解见文或直接用迭法求解醔例72008安徽文21设数列{an}满足a1=a,an+1=can+1c,n∈N*其中a醓c为实数且c≠0求数列{an}的通式解方法一因为an+11=c(an1)所a≠1时{an1}是首为a1比为c的等比数列所an1=(an1)cn1即an=(an1)cn1+1n=1时an=1满足式数列{an}的通式为an=(a1)cn1+1(n∈N*)方法由题设得n鄁2时,an1=c(an11)=c2(an21)=…=cn1(an1)=(a1)cn1所an=(a1)=cn1+1n=1时a1=a也满足式所{an}的通式为an=(a1)cn1+1(n∈N*)类型4的变式an+1=pan+f(n)解法思路通过构造新数列{bn}消去f(n)带来的差异例如面的类型5an+1=pan+qn其中p醓q均为常数pq(p1)(q1)≠0或an+1=pan+rqn其中p醓q醓r均为常数解法思路一般地要先在原递推式两边除qn+1得qqaqpqannnn111+⋅=++引入辅数列{bn}其中nnnqab=得qbqpbnn1+=即可转为类型3醔或直接将原递推式变形为nnnnqxapqxa⋅+=⋅+++(11其中)(1qpqpx≠−=则直接转为等比数列例82006全I理2212设数列{an}的前n的和...3,2,1,32231341=+×−=+nasnnn6求首a1通an醔111111111111442:123334124122,2(2)33333342,24(2),2244,42nnnnnnnnnnnnnnnnnnnnnasaanassaaaaaaaaa−−−−−−−===−+⇒=≥=−=−×+−−×+=++=+=+=⋅=−
解时时即所所所所例92009全II理19设数列{an}的前n的和24,1,11+==+nnnasas已知1设nnnaab21−=+证明数列{bn}是等比数列2求数列{an}的通式醔21111111121112212111212)13(414343)1(21243,21}2{43222322,3}{)1()2(2,3}{2)2(2244)24(2432523,24)1(:−−+−++++++++++⋅−=−=×−+===×=−===−=−−=+−+=−==−==++=+nnnnnnnnnnnnnnnnnnnnnnnnnnnnnannaaaaaaqbb，bbbaaaaaaaassaaab，aaaa所的等差数列差为是首为因数列于是所比中等比数列知由的等比数列比为是首为因数列即于是故解得由已知有解类型621apaqannn=+++其中pq均为常熟醔解法一待定系数法先把原递推式转为211()nnnnasatasa+++−=−其中s,t满足stpstq+==−解法特征根法对于由递推式21nnnapaqa++=+,1a=α,2a=β给出的数列{an}方程20xpxq−−=,做数列的特征方程醔若12,xx是特征方程的两个根12xx≠时数列{an}的通为1112nnnaAxBx−−=+其中A醓B由1a=α,2a=β决定即把1212,,,aaxx和n=1,2,入1112nnnaAxBx−−=+得到关于A醓B的方程组12xx=时数列的通为()11nnaABnAx−=+其中A醓B由1a=α,2a=β决定即把1212,,,aaxx和n=1,2入()11nnaABnAx−=+得到关于A醓B的方程组醔例102006福建文22已知数列{an}满足1a=1,2a=32132nnnaaa++=−nN∗∈醔1证明数列{}2nnaa+−是等比数列72求数列{an}的通式3若数列{bn}满足()121114441nnbbbbna−−−=+LnN∗∈证明{bn}是等差数列醔解1Q2132nnnaaa++=−∴2112()nnnnaaaa+++−=−Q1a=1,2a=3∴()21112nnnnnnaaaaaa++++−=−−nN∗∈∴{}2nnaa+−是21aa−=2为首2为比的等比数列醔222nnnaa+−=nN∗∈an=()1nnaa−−+()12nnaa−−−+L+()21aa−+1a=12n−+22n−+L+2+1=2n-1nN∗∈类型7递推式为Snna的关系式或Sn()nfa=解法思路这种类型一般利用na=()()11122nnnsnnassn−=−−≥LLLLLL或na=1nnss−−()nfa=()1nfa−−()2n≥消去na进行求解醔例11.2009湖理19已知数列{an}的前和Sn=-na-112n−+2n为整数nb=2nna求证数列{bn}是等差数列并求数列{an}的通式解在Sn=−na−112n−+2中n=1可得S1=-1a+1=1a2n≥时Sn-1=−1na−−212n−+2∴na=Sn−Sn-1=−na+1na−−112n−2na=1na−+112n−即2nna=12n−1na−+1又nb=2nna∴nb=1nb−+1即2n≥时nb-1nb−=1又1b=21a=1∴数列{bn}是首和差均为1的等差数列于是nb=n=2nna,∴na=2nn.例122008全II理20设数列{an}的前n和为Sn已知1a=a,1na+=Sn+3nnN∗∈,釙设nb=ns-3n求数列{bn}的通式釚若1na+鄁nanN∗∈求a的取值范围醔解釙依题意1ns+-ns=1na+=ns+3n即ns=2ns+3n由得1ns+-13n+=2ns-3n因所求通式为nb=ns-3n=a-312n−nN∗∈醔8釚由釙知ns=3n+a-312n−nN∗∈于是2n≥时na=ns-1ns−=3n+a-312n−-13n−-a-322n−=以×13n−+(a-3)22n−1nnaa+−=4×13n−+(a-3)22n−=22n−231232na−•+−,2n≥时,231232na−•+−鄁0⇔a鄁9醔又2a=1a+31a综所求的a的取值范围是[)9,−+∞醔类型8an+1=pan+an+b(p≠令,a≠代)解法思路这种类型一般利用待定系数法构造等比数列即()()11nnaxnypaxny++++=++已知递推式比较解出,xy而转为{}naxny++是比p为的等比数列醔例令3.以代代6山东文以以已知数列{an}中1a=12点()1,2nnnaa+−在直线yx=其中1,2,3n=L釙13nnnbaa+=−−求证数列{bn}是等比数列釚求数列{an}的通醔所{bn}是34−为首12为比的等比数列112211121111211111,2,233113,1,442241,1,(1)12211112.12nnnnnnnnnnnnnnnnnnnnnnaaanaaabaabaaananbaabaaaaaaaa+++++++++++++==+=−=−=−=−=−−=−−+++−−−∴===−−−−−−=−−Q解釙由已知得91121322111213131,4222311,22311,2231122,3112231111)(),222232.2nnnnnnnnnnnnnbaaaaaaaaaanan−+−+−=−×=−×∴−−=−×∴−−=−×−−=−×∴−−=−×∴−−−=−