数列通项公式的求法---函数不动点、方程特征根
利用函数的不动点(方程的特征根)
22bb,2bb,2221 若数列满足,且是方程的最小根,则xaxbxa,,,,0xxaxbx,,,,,,,,nnn,n14a4a
2xx,,,,,,,( ,,,nn1
2 已知数列x满足,求数列x的通项公式( xxxx,,,,241,1,,,,nnnnn,11
22xx,,,121xxx,,,241 解:令,则是其最小根,得,由题意知, x,,1x,0,,nn,1n
两边取对数,得log12log11xx,,,,,两边同时加1,得: ,,,,212nn,
log112log11xx,,,,,, ,,,,,,212nn,
故log11x,,是首项为log112x,,,公比为2的等比数列, ,,,,,,2n21
n21,nx,,21log112x,,,所以 , 故 ( ,,n2n
axb,axb,n10,02 若数列x满足xcadbc,,,,且( x,,,,,n,1n1cxd,cxd,1n
xx,,,,ac,,axb,nn,1,,, 2(1 若方程,,有两个相异实根,则 ( x,xacx,,,cxd,,,,nn,1
72x,n已知数列xxx,,,3x满足,求数列的通项公式( ,,,,nn,11nx,4n
xx,,11672x,nn,1,,,,1,2,,解:令,得为其两根,所以有, x,xx,,252x,4nn,1
,,x,1x,161n 所以数列是以为首项,以为公比的等比数列, ,2,,x,25x,21n,,
n,11x,16,,nx,,2 所以, 故 ( ,,2n,,n,1x,256,,,,n,,21,,5,,
121caxb,,,ad,,2(2 若方程有两个相等实根,且,则( x,,,,,,,xadxcxd,nn,1
31x,1nxxx,,,x 已知数列满足,求数列的通项公式( ,,,,nnn,11472x,n
11431x,1,,解:令,得为其根,所以 , x,x,,11547x,2xx,,nn,122
,,
14,,1 所以数列是以为首项,以为公差的等差数列, ,1,,115,,x,x,1n2,,2
1494,n 所以, 故 ( ,,,,n11x,,,n1528,nx,n2
22axc,axc,n,,,x, 3 若数列满足,若是方程的两个相异实根,则x0xa,,,,,,n,n12axf,2axf,n
2,,xx,,,,nn,1 ,,,xx,,,,nn,1,,
232x,19n 已知数列x满足,求数列x的通项公式( xx,,,,,,,nn,n11656x,n
211,,xx,,2nn,1,,32x,133解:令x,,得为其两根,所以, ,,,,,,,2,,65x,xx,,223nn,1,,
,,
11xx,,nn,133log2log,两边取对数,得, 33xx,,22nn,1
11,,x,x,1n,,33log1,所以数列是以为首项,以2为公比的等比数列, log,,33x,2x,21n,,
,,
1n,1x,2n631,,n,13x,log2, 所以 , 故 ( n,1n32x,2333,,n
2,,,,fx()0,(),,,fx()fx() 相关高考:已知函数,是方程的两个根,是 的fxxx()1,,,
fa()n,,,,,aan(12)导数(设,( a,1,1nn1,fa()n,,,(1)求的值;
a,,nln(12)bbn,,,,(2)已知对任意的正整数有,记(求数列的前项a,,nn,,nnna,,n和( Sn
,,15,,15,,,,解:(1)求根公式得, 22
2a,122n,fxx()21,, (2) ,,,,,,,,1,1a,,n121a,n222aaaaaa,,,,,,,212,,,,,,2nnnnnn,1 bb,,,,,lnlnlnln()2nn,1222aaaaaa,,,,,,,212,,,,,,nnnnnn,1
,,a51,1 ?数列是首项,公比为2的等比数列 , {}bq,b,,ln4lnn1a,2,1
nbq(1),51,n1 ? S,,,,4(21)lnn12,q
不动点法
pa,qpa,qnna,a,,如果数列满足下列条件:已知的值,且对于,都有(其中n,N{a}a,1,1nnn1ra,hra,hnn
hpx,qph,qrr,a,,p、q、r、h均为常数,且),那么,可作特征方程, x,,0,1rrx,h
,,1当特征方程有且仅有一根时,则是等差数列; x,,0ax,n0,,
,,ax,n1当特征方程有两个相异的根、时,则是等比数列( xx,,12ax,n2,,
2124a,n例、已知数列满足aa,,,4,求数列的通项公式( {}a{}an,11nn41a,n
2124x,2124x,2420240xx,,,解:令,得,则是函数的两个不动点(因xx,,23,x,fx(),1241x,41x,为
2124a,n,2,,aaaaaa,,,,,,,24121242(41)1326213a,2nnnnnn,1n,,,,(所以数列是以,,2124a,aaaaa,,,,,,321243(41)92793a,3nnnnnn,1n,,,341a,n
a,2a,242,13113n,1n1,2(),,2为首项,以为公比的等比数列,故,则a,,( 3n139a,39a,,343,1n1n,2()19
2124x,2124x,评注:本题解题的关键是先求出函数的不动点,即方程的两个根fx(),x,41x,41x,
,,aa,,2213a,2nn,1n,,xx,,23,,进而可推出,从而可知数列为等比数列,再求出数列,,12aa,,393a,3nn,1n,,,,a,2n的通项公式,最后求出数列的通项公式( {}a,,na,3n,,
72a,n例、已知数列满足,求数列的通项公式( aa,,,2{}a{}an,11nn23a,n
72x,31x,22420xx,,,解:令,得,则是函数的不动点( x,1x,fx(),23x,47x,
7255aa,,nn因为,所以 a,,,,11,1n2323aa,,nn
35a,na,2312212n22, ,,,,,,,(1)aaaaa,,,,,155515115,1nnnnn
,,121112所以数列是以为首项,以为公差的等差数列,则,,,n,故,,11(1),,5a,,121a,15a,1n1n,,
28n,( a,n23n,
31x,72x,评注:本题解题的关键是先求出函数的不动点,即方程的根,进而可推出x,1fx(),x,47x,23x,
,,,,11211,,,从而可知数列为等差数列,再求出数列的通项公式,最后求出数,,,,aa,,115a,a,11nn,1nn,,,,列的通项公式( {}an
特征根法
2对于由递推公式,给出的数列,方程,叫做数列a,pa,qa,,ax,px,q,0a,,,a,,n,2n,1nn12
的特征方程(若是特征方程的两个根, ,,ax,xn12
nn当时,数列的通项为aAxBx,,,其中A,B由决定(即把和,,ax,xa,,,a,,a,a,x,xnn1212121212
nnn,1,2,代入aAxBx,,,得到关于A、B的方程组); n12
n当时,数列的通项为aABnx,,(),其中A,B由决定(即把,,ax,xa,,,a,,a,a,x,xn112121212n
nn,1,2和,代入aABnx,,(),得到关于A、B的方程组)( 1n
例、已知数列满足,求数列的通项公式( aaanaa,,,,,3(2)1,{}a{}annnn,,1112n
2,,,,,310解:aaan,,,3(2)的相应特征方程为,解之求特征根是nnn,,11
nn,,,,3535,,,,3535,,,,,,所以(由初始值aa,,1,得方程组acc,,,,,,121212n,,,,2222,,,,
,,525,3535,,11c,1()(),,cc,,112,,522求得 ,,3535,,525,,,221()(),,ccc,122,,,225,
,,,,5253552535nna,,从而()()( n5252
c,c评注:本题解题的关键是先求出特征方程的根(再由初始值确定出,从而可得数列的通项公式( {}a12n
,nN,在数列{}中,,当,,求通项公式. a,5a,6aaaa,,1,a,2n,2n,1nnn12
2,,,,,560解:的相应特征方程为,解之求特征根是,所以a,5a,6a,,,,23,n,2n,1n12
5,c,,1,,,,123cc,,212nn(由初始值,得方程组求得 acc,,23a,,1,a,2,,1212n4249,,cc,12,c,2,3,54nnnn,,11从而( a,,,,,234352n23
待定系数法:原式可化为: a,,a,(5,,)(a,,a)n,2n,1n,1n
比较系数得=-3或=-2, ,,
不妨取,=-2.?式可化为: a,2a,3(a,2a)n,2n,1n,1n则是一个等比数列,首项=2-2(-1)=4,公比为3. {a,2a}a,2an,1n21
n,1n,1n,1?a,2a,4,3.有:a,4,3,5,2. n,1nn