1、 25 ()()012abcbcaxbc+ + na 11,1/,1,2.nnaaa a an+=+=“ (I) X na Ki Ov , , p A = limnna(| A aV U ); (II) ! ,1,2,nnbaAn= = “ : 1nb+= /( ( );nnbAb A+ (III) |1/2 1,2,nnbn=“对 , pa |S . (2004 M I S k5 225 ) 点评 5 L BQs Tf () ( 1)/f xax x= + /w 5 . = # Ka Taw a T (B ,Eab E ).58C K XaZ Xa/ X#B , X .1 I a K Q B
2、 ,E , I I 3 2 Ms5 %5 ? .(sV ) / A+ Ms 8 g IBQsTw 5 ,LB BQs Tf () ( )/( )( ),f xaxbcxdadbc= +/w 5 =d . 1一次分式递推数列的迭代周期 BQs Tf () ( ),ax bf xadbccx d+=+“ “B= abAcd=,726 T2() ( ()f xffx=22()( )()()a bc x ab bdac cd x bc d+=+“ “ A Z 222.ab bdabcAac cd bc d+=+B , T ()nf x “ nA . ,BQs Tf nQ p A n QZ .+Y ,0
3、0ntAt=(0),t L ( () )ff fx“x= , f . 例1(2005 M Ic 2 c S 55 )X na 1130,31nnnaaaa+=+(n *)N ,520a = ( ) (A)0, (B) 3 , (C) 3 , (D) 3/2. 解 *1130, ( )31nnnaaa nNa+= +, 1213 33,131aaa =+2323 333,313(3)1aaa = =+3433 330,313(3)1aaa = =+ VB , na 3T = .V7 ,20 6 3 2 23.aa a+=# B. 2一次分式递推数列的无穷数列与有穷数列 BQs Tf ()f x
4、= (ax + )/( )bcxd+ (),ad bc / na k ,# k H , , ,abcd1 ,7 O S1a 11. 例2 (2005 M Ic y c S 225 ) X na 11,11/nnaaa a+=+. a | H , . 1a = H , k :1,2,3/2,5/3,“ ; 1/2a = H , k :1/2,1,0. (1) p a | H ,40;a = (2) ! nb 111, 1 /( 1)nnbb b+= = (n *)N , p a | nb B , VB k na ; (3) 2/3 2( 4),nan若取 3A =则 )8C 5 .7 I a1
5、a TM , I I 3 B ,E %15? . % V?C5 - 2004 M I S k5 225 B . 5利用一次分式函数构造递推数列问题 M I5 ,B k5 . aBt MZE ,I/ k5 , T vm . / BQs Tw /BB 3 ?) 5 . 例6 Xf () ,21xfxx=+ na*111,() ,2)2006nnaafanNn= = （ . (1) p na Y T ; (2) 71/nnnbaa= , p nb KvaKl#M n . L I 5T ,1 p TB = ,1a C ,g W = “ ,ZE , “/ dBt 5 , ) il .“S; 4 6 =
6、? , = , Sa ?ZaE =u/T L$ ; 3 . ID 1 . 5 5 . 2.1993 2 .w 1/nnapqa+=+ Y T# . Y ,2001.13. 3 . ZE ps TB nx 11()/()nnax b cx d= +Y T . =Sv (1 S ).2005.3. 例谈函数值域的求法 y u ? pf B1 = ,9 f B 4 .Bf lE5 ,f 9 .f pE 2“ , # M z , XZE , pA-%4f Vr T+ ,|ME , 2oMYp ./ + pf ZE . 1观察法 Z a a d+ ,#f ma a e9 aw , 4 ?Btef .YVf l# Ts , f ,4 pf .
