二阶递推数列整理

2021-03-21 01:08 作者:Cuprate 0人读过 | 我要投稿

本章前置芝士：数列、不动点、特征方程

在教辅中看到几道不错的题目，先暂且记录下来

关于二阶递推式变式转换（与下文例题无关）

转换1：

$a_%7Bn%2B2%7D%3Dp*a_%7Bn%2B1%7D%2Bq*a_n%5Crightarrow%20a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n%3Da*b%5En(p%2Cq%5Cneq0)$

证明：

$%5Cbecause%20p*a_%7Bn%2B1%7D%3Da_%7Bn%2B2%7D-q*a_n%2Cp*a_%7Bn%2B2%7D%3Da_%7Bn%2B3%7D-q*a_%7Bn%2B1%7D$

$%5Ctherefore%20p*a_%7Bn%2B1%7D*(a_%7Bn%2B3%7D-q*a_%7Bn%2B1%7D)%3Dp*a_%7Bn%2B2%7D*(a_%7Bn%2B2%7D-q*a_n)$

$%5Cbecause%20p%5Cneq0$

$%5Ctherefore$ 整 $a_%7Bn%2B2%7D%5E2-a_%7Bn%2B3%7D*a_%7Bn%2B1%7D%3D(-q)(a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n)$

$%5Ctherefore%20a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n%3D(-q)%5E%7Bn-1%7D(a_2%5E2-a_3*a_1)$

$%5Cquad%3D%5Cfrac%7B1%7D%7Bq%7D(q*a_1%5E2%2Bp*a_2a_1-a_2%5E2)(-q)%5En$

转换2： $a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n%3Da*b%5En%5Crightarrow%20a_%7Bn%2B2%7D%3Dp*a_%7Bn%2B1%7D%2Bq*a_n%20(%5Cforall%20n%5Cin%20N%5E*%2Ca_n%5Cneq0)$

证明：

$%5Cbecause%20a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n%3Da*b%5En%2Ca_%7Bn%2B2%7D%5E2-a_%7Bn%2B3%7D*a_%7Bn%2B1%7D%3Da*b%5E%7Bn%2B1%7D$

$%5Ctherefore%20a_%7Bn%2B2%7D%5E2-a_%7Bn%2B3%7D*a_%7Bn%2B1%7D%3Db(a_%7Bn%2B1%7D%5E2-a_%7Bn%2B2%7D*a_n)$

$%5Cquad%20a_%7Bn%2B2%7D*(a_%7Bn%2B2%7D%2Bb*a_%7Bn%7D)%3Da_%7Bn%2B1%7D*(a_%7Bn%2B3%7D%2Bb*a_%7Bn%2B1%7D)$

$%5Cbecause%20%5Cforall%20n%5Cin%20N%5E*%2Ca_n%5Cneq0$

$%5Ctherefore%5Cfrac%7Ba_%7Bn%2B3%7D%2Bb*a_%7Bn%2B1%7D%7D%7Ba_%7Bn%2B2%7D%7D%3D%5Cfrac%7Ba_%7Bn%2B2%7D%2Bb*a_n%7D%7Ba_%7Bn%2B1%7D%7D$

$%5Ctherefore%5Cfrac%7Ba_%7Bn%2B2%7D%2Bb*a_n%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cfrac%7Ba_3%2Bb*a_1%7D%7Ba_2%7D$

又 $%5Cbecause%20a_3%3D%5Cfrac%7Ba_2%5E2-ab%7D%7Ba_1%7D$

$%5Ctherefore%5Cfrac%7Ba_3%2Bb*a_1%7D%7Ba_2%7D%3D%5Cfrac%7Ba_2%5E2%2Bb*a_1%5E2-ab%7D%7Ba_1a_2%7D$

$%5Ctherefore%20a_%7Bn%2B2%7D%3D%5Cfrac%7Ba_2%5E2%2Bb*a_1%5E2-ab%7D%7Ba_1a_2%7Da_%7Bn%2B1%7D-ba_n$

例题1： $a_1%3Da_2%3D1%2Ca_%7Bn%2B2%7D%3D3a_%7Bn%2B1%7D%2B18a_n-2*5%5En$ ，求 $a_n$

例题2： $a_1%3D1%2Ca_2%3D2%2Ca_%7Bn%2B2%7D%3D7a_%7Bn%2B1%7D-a_n$ ，证明 $a_na_%7Bn%2B1%7D-1$ 是完全平方数

例题3： $a_1%3Da_2%3Da_3%3D1%2Ca_%7Bn%2B3%7D%3D%5Cfrac%7Ba_%7Bn%2B2%7D*a_%7Bn%2B1%7D%2Bk%7D%7Ba_%7Bn%7D%7D(k%3E0)$

$(1)$ 证 $2a_%7B2n%7D%3Da_%7B2n-1%7D%2Ba_%7B2n%2B1%7D%2C(k%2B2)a_%7B2n%2B1%7D%3Da_%7B2n%2B2%7D%2Ba_%7B2n%7D$

$(2)$ 证 $(k%2B2)a_%7B2n%2B1%7D%5E2%2Bk%3D2a_%7B2n%2B2%7D*a_%7B2n%7D%2C2a_%7B2n%7D%5E2%2Bk%3D(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D$

$(3)$ 求 $a_n$

例题1解答：

先忽略 $2*5%5En$ 项，用特征方程 $x%5E2-3x-18%3D0$ 得特征根 $%5Clambda_1%3D6%2C%5Clambda_2%3D-3$

将原递推式转化为 $a_%7Bn%2B2%7D%2B3a_%7Bn%2B1%7D%2BA*5%5E%7Bn%2B1%7D%3D6(a_%7Bn%2B1%7D%2B3a_n%2BA*5%5En)$

待定系数得 $A%3D-2$

又 $%5Cbecause%20a_2%2B3*a_1-10%3D-6$

$%5Ctherefore%20a_%7Bn%2B1%7D%2B3a_n-2*5%5En%3D-6%5En$

同理，将递推式转化为 $a_%7Bn%2B1%7D%2BB_1*5%5E%7Bn%2B1%7D%2BB_2*6%5E%7Bn%2B1%7D%3D-3(a_n%2BB_1*5%5En%2BB_2*6%5En)$

待定系数得 $B_!%3D-%5Cfrac%7B1%7D%7B4%7D%2CB_2%3D%5Cfrac%7B1%7D%7B9%7D$

又 $%5Cbecause%20a_1-%5Cfrac%7B5%7D%7B4%7D%2B%5Cfrac%7B6%7D%7B9%7D%3D%5Cfrac%7B5%7D%7B12%7D$

$%5Ctherefore%20a_n%3D%5Cfrac%7B5%5En%7D%7B4%7D-%5Cfrac%7B6%5En%7D%7B9%7D-%5Cfrac%7B5(-3)%5En%7D%7B36%7D$

例题2解答：

由 $a_%7Bn%2B2%7D%3D7a_%7Bn%2B1%7D-a_n$ 得 $(a_%7Bn%2B2%7D-a_n)(a_%7Bn%2B2%7D-7a_%7Bn%2B1%7D%2Ba_n)%3D0$

整理得 $a_%7Bn%2B2%7D%5E2-7a_%7Bn%2B2%7Da_%7Bn%2B1%7D%3Da_n%5E2-7a_na_%7Bn%2B1%7D$

等号两边同加 $a_%7Bn%2B1%7D%5E2$ 得 $a_%7Bn%2B2%7D%5E2%2Ba_%7Bn%2B1%7D%5E2-7a_%7Bn%2B2%7Da_%7Bn%2B1%7D%3Da_%7Bn%2B1%7D%5E2%2Ba_n%5E2-7a_na_%7Bn%2B1%7D$

$%5Ctherefore%20a_%7Bn%2B1%7D%5E2%2Ba_n%5E2-7a_na_%7Bn%2B1%7D%3Da_2%5E2%2Ba_1%5E2-7a_2a_1%3D-9$

$%5Ctherefore%20(a_%7Bn%2B1%7D%2Ba_n)%5E2%3D9(a_%7Bn%2B1%7Da_n-1)$ ， $a_%7Bn%2B1%7Da_n-1$ 是完全平方数

例题3简答：

$(1)a_4%3D%5Cfrac%7Ba_3*a_2%2Bk%7D%7Ba_1%7D%3Dk%2B1$

$%5Cbecause%20a_%7Bn%2B3%7D*a_n-a_%7Bn%2B2%7D*a_%7Bn%2B1%7D%3Dk%3Da_%7Bn%2B4%7D*a_%7Bn%2B1%7D-a_%7Bn%2B3%7D*a_%7Bn%2B2%7D$

$%5Ctherefore%20a_%7Bn%2B1%7D*(a_%7Bn%2B4%7D%2Ba_%7Bn%2B2%7D)%3Da_%7Bn%2B3%7D*(a_%7Bn%2B2%7D%2Ba_n)$

$%5Cquad%5Cfrac%7Ba_%7Bn%2B4%7D%2Ba_%7Bn%2B2%7D%7D%7Ba_%7Bn%2B3%7D%7D%3D%5Cfrac%7Ba_%7Bn%2B2%7D%2Ba_n%7D%7Ba_%7Bn%2B1%7D%7D(n%5Cin%20N%5E*)$

$%5Ctherefore%5Cfrac%7Ba_%7B2n%2B1%7D%2Ba_%7B2n-1%7D%7D%7Ba_%7B2n%7D%7D%3D%5Cfrac%7Ba_%7B2n-1%7D%2Ba_%7B2n-3%7D%7D%7Ba_%7B2n-2%7D%7D%3D...%3D%5Cfrac%7Ba_3%2Ba_1%7D%7Ba_2%7D%3D2$

$%5Cquad%5Cfrac%7Ba_%7B2n%2B2%7D%2Ba_%7B2n%7D%7D%7Ba_%7B2n%2B1%7D%7D%3D%5Cfrac%7Ba_%7B2n%7D%2Ba_%7B2n-2%7D%7D%7Ba_%7B2n-1%7D%7D%3D...%3D%5Cfrac%7Ba_4%2Ba_2%7D%7Ba_3%7D%3Dk%2B2$

$%5Ctherefore2a_%7B2n%7D%3Da_%7B2n-1%7D%2Ba_%7B2n%2B1%7D%2C(k%2B2)a_%7B2n%2B1%7D%3Da_%7B2n%2B2%7D%2Ba_%7B2n%7D%20(n%5Cin%20N%5E*)$

$(2)2(k%2B2)a_%7B2n%7D*a_%7B2n%2B1%7D%3D2(k%2B2)a_%7B2n%2B1%7D*a_%7B2n%7D$

$%5Cbecause2a_%7B2n%7D%3Da_%7B2n-1%7D%2Ba_%7B2n%2B1%7D%2C(k%2B2)a_%7B2n%2B1%7D%3Da_%7B2n%2B2%7D%2Ba_%7B2n%7D$

$%5Ctherefore(k%2B2)a_%7B2n%2B1%7D*(a_%7B2n%2B1%7D%2Ba_%7B2n-1%7D)%3D2a_%7B2n%7D*(a_%7B2n%2B2%7D*a_%7B2n%7D)$

整理得 $2a_%7B2n%7D%5E2-(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D%3D(k%2B2)a_%7B2n%2B1%7D%5E2-a_%7B2n%2B2%7D*a_%7B2n%7D(n%5Cin%20N%5E*)$

同理，可由 $2(k%2B2)a_%7B2n%2B1%7D*a_%7B2n%2B2%7D%3D2(k%2B2)a_%7B2n%2B2%7D*a_%7B2n%2B1%7D$ 推得

$2a_%7B2n%2B2%7D%5E2-(k%2B2)a_%7B2n%2B3%7D*a_%7B2n%2B1%7D%3D(k%2B2)a_%7B2n%2B1%7D%5E2-a_%7B2n%2B2%7D*a_%7B2n%7D$

$%5Ctherefore2a_%7B2n%2B2%7D%5E2-(k%2B2)a_%7B2n%2B3%7D*a_%7B2n%2B1%7D%3D2a_%7B2n%7D%5E2-(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D$

$%5Ctherefore2a_%7B2n%7D%5E2-(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D%3D2a_%7B2n-2%7D%5E2-(k%2B2)a_%7B2n-1%7D*a_%7B2n-3%7D%3D...%3D2a_2%5E2-(k%2B2)a_3*a_1%3D-k$

$%5Cquad(k%2B2)a_%7B2n%2B1%7D%5E2-a_%7B2n%2B2%7D*a_%7B2n%7D%3D2a_%7B2n%7D%5E2-(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D%3D-k$

$%5Ctherefore(k%2B2)a_%7B2n%2B1%7D%5E2%2Bk%3D2a_%7B2n%2B2%7D*a_%7B2n%7D%2C2a_%7B2n%7D%5E2%2Bk%3D(k%2B2)a_%7B2n%2B1%7D*a_%7B2n-1%7D%20(n%5Cin%20N%5E*)$

$(3)%5Cbecause%202a_%7B2n%7D%3Da_%7B2n-1%7D%2Ba_%7B2n%2B1%7D%2C(k%2B2)a_%7B2n%2B1%7D%3Da_%7B2n%2B2%7D%2Ba_%7B2n%7D%20(n%5Cin%20N%5E*)$

$%5Ctherefore2(k%2B2)a_%7B2n%2B2%7D%3D(k%2B2)a_%7B2n%2B1%7D%2B(k%2B2)a_%7B2n%2B3%7D%3Da_%7B2n%7D%2B2a_%7B2n%2B2%7D%2Ba_%7B2n%2B4%7D$

$%5Cquad(2k%2B2)a_%7B2n%2B2%7D%3Da_%7B2n%7D%2Ba_%7B2n%2B4%7D$

同理， $(2k%2B2)a_%7B2n%2B1%7D%3Da_%7B2n-1%7D%2Ba_%7B2n%2B3%7D$

$%5Ctherefore%20(2k%2B2)a_%7Bn%2B2%7D%3Da_%7Bn%2B4%7D%2Ba_n%20(n%5Cin%20N%5E*)$

不妨设 $b_n%3Da_%7B2n-1%7D(n%5Cin%20N%5E*)$

$b_%7Bn%2B2%7D-(2k%2B2)b_%7Bn%2B1%7D%2Bb_n%3D0$ 对应特征方程 $%5Clambda%5E2-(2k%2B2)%5Clambda%2B1%3D0$

$%5CDelta%3D(2k%2B2)%5E2-4%3D4k%5E2%2B8k$

$%5Cbecause%20k%3E0$

$%5Ctherefore%5CDelta%3E0$

特征根 $%5Clambda_1%3Dk%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D%2C%5Clambda_2%3Dk%2B1-%5Csqrt%7Bk%5E2%2B2k%7D%2C%5Clambda_1%3E%5Clambda_2%3E0$

$%5Ctherefore%20b_n%3Dc_1*(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5En%2Bc_2*(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)%5En$

由 $b_1%3Db_2%3D1$ 联立方程

$%5Cbegin%7Bcases%7D(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)*c_1%2B(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)*c_2%3D1%20%5C%5C(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5E2*c_1%2B(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)%5E2*c_2%3D1%5Cend%7Bcases%7D$

解得 $%5Cbegin%7Bcases%7Dc_1%3D%5Cfrac%7B1%7D%7B(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)*(k%2B2%2B%5Csqrt%7Bk%5E2%2B2k%7D)%7D%20%5C%5Cc_2%3D%5Cfrac%7B(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5E2%7D%7Bk%2B2%2B%5Csqrt%7Bk%5E2%2B2k%7D%7D%5Cend%7Bcases%7D$

$%5Ctherefore%20b_n%3D%5Cfrac%7B1%7D%7Bk%2B2%2B%5Csqrt%7Bk%5E2%2B2k%7D%7D%5B(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5E%7Bn-1%7D%2B(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)%5E%7Bn-2%7D%5D(n%5Cin%20N%5E*)$

同理，可设 $d_n%3Da_%7B2n%7D(n%5Cin%20N%5E*)$

$%5Ctherefore%20a_n%3D%5Cbegin%7Bcases%7D%5Cfrac%7B1%7D%7Bk%2B2%2B%5Csqrt%7Bk%5E2%2B2k%7D%7D%5B(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5E%5Cfrac%7Bn-1%7D%7B2%7D%2B(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)%5E%5Cfrac%7Bn-3%7D%7B2%7D%5D%5Cquad%20n%3D2t-1%5C%5C%5Cfrac%7B1%7D%7B2%7D%5B(k%2B1%2B%5Csqrt%7Bk%5E2%2B2k%7D)%5E%7B%5Cfrac%7B1%7D%7B2%7Dn-1%7D%2B(k%2B1-%5Csqrt%7Bk%5E2%2B2k%7D)%5E%7B%5Cfrac%7B1%7D%7B2%7Dn-1%7D%5D%5Cqquad%5Cqquad%5C%20%20n%3D2t%5Cend%7Bcases%7D%5Cqquad%20t%5Cin%20N%5E*$