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1、 at +b用不动点法求递推数列(a +c 0)的通项2t =n+12nct + dn储炳南（安徽省岳西中学 246600）1通项的求法at +bt =n+1为了求出递推数列的通项，我们先给出如下两个定义：nct + dn= f (t )( )，则称 f x 为数列t 的特征函数.定义 1：若数列t 满足tnnn+1n(x)( )f x( )f x定义 2:方程 f=x 称为函数的不动点方程，其根称为函数的不动点.at +bt =n+1下面分两种情况给出递推数列n通项的求解通法.ct + dn(1)当 c=0,时,at +babt =n+1tt= +由n,ct + dn+1 dndnabt =
2、 k t + c记 = k , = c ，则有（k0）, t -n+1ndd数列t 的特征函数为 f(x)=kx+c,nccc由 kx+c=xx=,则= k(t -)t= k t + c1- k1- k1- kn +1nn+1nc数列t -是公比为 k 的等比数列,1- kncccct -= (t -) k t =+ (t -) k .n-1n-11- k(2)当 c0 时,1- k1- k1- kn1n1a x + bf(x)数列 的特征函数为：t=c x + dn1 a x +bc x + d= x2由 cx + (d - a)x - b = 0设方程cx + (d - a)x - b =
3、 0 的两根为 x ,x ,则有:212cx2+ (d - a)x -b = 0,cx + (d - a)x - b = 022121b = cx + (d - a)x (1)211b = cx2 + (d - a)x (2)22t - xt - x又设= (其中,nN ,k 为待定常数).*kn+11n1t - xt - xn+12n2a t + b- xnt - xt - xt + dt - xc1由= k = kn+11n11nnt xa t + b- xt - x-t- xn+12n2n2nc t + d2nat + b - cx t - dxt - x= k (3)1n1 n1nat
4、 + b - cx t - dxt - xn2 n2n2将(1)、（2）式代入(3)式得:at + cx - cx t - ax t - x21= k n1 n1n1at + cx - cx t - axt - x22n2 n2n2(a - cx )(t - x )t - xa cx-= k k =1n1n11(a - cx )(t - x )t - xa - cx2n2n22t - x- cxa - cxa数列是公比为(易证 0 )的等比数列.1n11t - xa - cxa - cxn222t - x t- x a - cx n-1=1111nt - x t- xa - cx2122n2
5、t - x -a cx n-1x - x 111- xa - cx12t =.t122nt x- -a cxn-11-111t - xa - cx1222应用举例2a +1例 1：已知数列a 中，a =2， a =，求a 的通项。nn3n1+1n2x +1解：因为a 的特征函数为： f (x) =,3n2x +1由 f (x) = x x = 1,32a +12 a = a -1 = (a -1)n33n+1n+1n2数列a -1是公比为 的等比数列,3n22a -1=(a -1) ( ) a =1+( ) .n-1n-133n1n4a - 2例 2 已知数列a 中，a =3，n=，求a 的通
6、项。nan+11n+1an4x - 2解：因为a 的特征函数为： f (x) =,x+1n4x - 2由 f (x) = x x - 3x + 2 = 0 x = 1, x = 22x +1124a - 2-1- 2na -1a -1+1a -1a设= k = k = k n+1nnn4a - 2a - 2n+1a - 2- 2annna +1n3a - 3a -13 (a -1)a -1= k nnnn2a - 4a - 22 (a - 2)a - 2nnnn3a -13-1- 2aa k = 即= ,n+1n2- 2 2an+1n3 -13a数列是公比为 的等比数列.2na - 2na -
7、1 a -1 n-13= n1a - 2 a - 2 2 n1a -132 - 23 n-1n-2n-1a =3,1= 2 a =. na- 22 2 - 3n-2n-1nn1+ a例 3 已知数列a 中，a =2，1=，求a 的通项。nann1- ann+11+ x1- x解：因为a 的特征函数为： f (x) =,n1+ x由 f (x) = x x +1 = 0 x = i, x = -i21- x121+ a1- a1+ a1- a- i+ ina - ia - ia - i设= k k= n+1nnnna + in+1a + i+ iannn1+ a - i + a ia - i1+
8、 (a - i)a - ii= k = k nnnnn1+ a + i - a ia + i1- i (a + i)a + innnnn1+ i-1+ i -a iaaik =即=i a i,n+1n+1n1- i+1-+in - a i1+ i数列是公比为的等比数列.na + in1- ia - i a - i 1+ in-1=1na + i a + i 1- i n1a - i2 - i 1+ ia i-2 - in-1()a =2,1=in-1nna i+ i 2 + i 1- i+2 +iann(2 - i)i -1+ 2in =.a2 + i - (2 - i)inn-14 1例 4
9、 已知数列a 的前 n 项和为 S ， a = ， S = n a - n(n -1)，求a 的通22n1nnnn项。解： S = n a - n(n -1)2nn S = (n +1) a - (n +1)n 2n+1n+1-得： a = (n +1) a - n a - (n +1)n + n(n -1)22n+1n+1nn2 (n + 2)a = na + 2 a =a +nn + 2n + 2+1nn+1nn2因为a 的特征函数为： f (x) =x +,n+ 2n + 2nn2由 f (x) =x += x x=1.n + 2n + 2设 a -1 = b a = b +1， a = b +1nnnnn+1n+1n2将代入得：b +1 =(b +1) +n + 2n + 2n+1nnbn b =b n=n+1n + 2n + 2n+1bn12b b bb b4b = L ， = -1 = -b a23nn1b b bb11123n-11 1 2 3n -1n +11b = - L = -2 3 4 5n(n +1)n1 a = b +1 = 1-。n(n +1)nn5