高中必學(xué)數(shù)列構(gòu)造

2023-07-03 20:03 作者:water_不是水 0人讀過(guò) | 我要投稿

1. $a_%7Bn%2B1%7D%3Dma_n%2Bm%5En$

兩邊同時(shí)除以 $ma_%7Bn%2B1%7D$ ，得到 $%5Cfrac%7Ba_%7Bn%2B1%7D%20%7D%7Bm%5E%7Bn%2B1%7D%7D%20-%5Cfrac%7Ba_n%7D%7Bm%5En%7D%3D%5Cfrac%7B1%7D%7Bm%7D%20$

記 $b_n%3D%5Cfrac%7Ba_n%7D%7Bm%5En%7D$ ，由等差數(shù)列公式可得

$%7B%5Ccolor%7BRed%7D%20%7Ba_n%3Dm%5E%7Bn-1%7D%5Cleft%20(%20r-1%20%5Cright%20)%2Bn%5Ccdot%20m%5E%7Bn-1%7D%7D%20%7D%20$

2. $a_%7Bn%2B1%7D%3Dma_n%2Bm%5En%2Bt$

待定系數(shù)法可列 $%5Cleft%20(%20a_%7Bn%2B1%7D-c%20%5Cright%20)-m%5Cleft%20(%20a_n-c%20%5Cright%20)%20%20%3Dm%5En$ ，解得 $c%3D%5Cfrac%7Bt%7D%7Bm%2B1%7D%20$ .

兩邊同時(shí)除以 $m%5E%7Bn%2B1%7D$ ，構(gòu)建 $b_n%3D%5Cfrac%7Ba_n-c%7D%7Bm%5En%7D$ 為等差數(shù)列。

3. $a_%7Bn%2B1%7D%3D%5Cfrac%7Baa_n%2Bb%7D%7Bca_n%2Bd%7D$

(1)寫(xiě)出 $x%3D%5Cfrac%7Bax%2Bb%7D%7Bcx%2Bd%7D$ ,解得 $x_1%2Cx_2$ .

(2)若 $x_1%5Cne%20x_2$ ，則 $%5Cleft%20%5C%7B%20%5Cfrac%7Ba_n-x_1%7D%7Ba_n-x_2%7D%20%20%5Cright%20%5C%7D%20$ 為等比數(shù)列。

若 $x_1%3Dx_2%3Dx$ ，則 $%5Cleft%20%5C%7B%20%5Cfrac%7B1%7D%7Ba_n-x%7D%20%20%5Cright%20%5C%7D$ 為等差數(shù)列。

若無(wú)實(shí)根，則數(shù)列有周期。

4. $a_%7Bn%2B1%7D%3Dca_n%2Bd%20%20%5Cleft%20(%20c%5Cne1%2Ccd%5Cne%200%20%20%5Cright%20)%20$

由待定系數(shù)法化為 $a_%7Bn%2B1%7D%2B%5Cfrac%7Bd%7D%7Bc-1%7D%20%3Dc%5Cleft%20(%20a_n%2B%5Cfrac%7Bd%7D%7Bc-1%7D%20%20%5Cright%20)%20$

則 $%5Cleft%20%5C%7B%20a_n%2B%5Cfrac%7Bd%7D%7Bc-1%7D%20%20%5Cright%20%5C%7D%20$ 為等比數(shù)列。

5. $a_%7Bn%2B1%7D%3Dma_n%2Br%5En$

(1)待定系數(shù)法有 $a_%7Bn%2B1%7D%2B%5Cfrac%7Br%5E%7Bn%2B1%7D%7D%7Bm-r%7D%3Dm%5Cleft%20(%20a_%7Bn%7D%2B%5Cfrac%7Br%5E%7Bn%7D%7D%7Bm-r%7D%20%5Cright%20)%20$

則 $%5Cleft%20%5C%7B%20a_n%2B%5Cfrac%7Br%5En%7D%7Bm-r%7D%20%20%5Cright%20%5C%7D$ 為等比數(shù)列。

(2)兩邊同時(shí)除以 $r%5E%7Bn%2B1%7D$ 得到 $%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Br%5E%7Bn%2B1%7D%7D%20%3D%5Cfrac%7Bm%7D%7Br%7D%5Ccdot%20%5Cfrac%7Ba_n%7D%7Br%5En%7D%20%20%2B%5Cfrac%7B1%7D%7Br%7D%20$

由待定系數(shù)法有: $%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Br%5E%7Bn%2B1%7D%7D%20%2B%5Cfrac%7B1%7D%7Bm-r%7D%3D%5Cfrac%7Bm%7D%7Br%7D%20%5Cleft%20(%20%5Cfrac%7Ba_n%7D%7Br%7D%2B%5Cfrac%7B1%7D%7Bm-r%7D%20%20%20%5Cright%20)%20$

則 $%5Cleft%20%5C%7B%20%5Cfrac%7Ba_n%7D%7Br%5E%7Bn%7D%7D%2B%5Cfrac%7B1%7D%7Bm-r%7D%20%20%5Cright%20%5C%7D$ 是等比數(shù)列。

6. $a_%7Bn%2B1%7D%3Dca_n%2Bpn%2Br$

由待定系數(shù)法有 $a_%7Bn%2B1%7D%2BA%5Cleft%20(%20n%2B1%20%5Cright%20)%20%2BB%3DC%5Cleft%20(%20a_n%2BAn%2BB%20%5Cright%20)%20$ .

其中 $A%3D%5Cfrac%7Bp%7D%7Bc-1%7D%20%2CB%3D%5Cfrac%7Bp%2Bcr-r%7D%7B%5Cleft%20(%20c-1%20%5Cright%20)%5E2%20%7D%20$ .

7. $a_%7Bn%2B1%7D%3Dma_n%2Bk%5Ccdot%20r%5En%2Bp$

(1)累加法

兩邊同時(shí)除以 $m%5E%7Bn%2B1%7D$ 得到 $%5Cfrac%7Ba_%7Bn%2B1%7D%7D%7Bm%5E%7Bn%2B1%7D%7D%20%3D%5Cfrac%7Ba_n%7D%7Bm%5En%7D%2B%5Cfrac%7Bk%7D%7Bm%7D%5Cleft%20(%5Cfrac%7Br%7D%7Bm%7D%20%20%20%5Cright%20)%5En%20$

記 $b_n%3D%5Cfrac%7Ba_n%7D%7Bm%5En%7D%20$ ，則 $b_%7Bn%2B1%7D-b_n%3D%5Cfrac%7Bk%7D%7Bm%7D%20%5Cleft%20(%5Cfrac%7Br%7D%7Bm%7D%20%20%20%5Cright%20)%20%5En$ .

(2)待定系數(shù)法

可化為 $a_%7Bn%2B1%7D%2B%5Cfrac%7Bk%7D%7Bm-r%7D%20%5Ccdot%20r%5E%7Bn%2B1%7D%3Dm%5Cleft%20(%20%20a_%7Bn%7D%2B%5Cfrac%7Bk%7D%7Bm-r%7D%20%5Ccdot%20r%5E%7Bn%7D%5Cright%20)%20$

8. $a_%7Bn%2B1%7D%3Dma_n%2Bk%5Ccdot%20r%5En%2Bp$

由待定系數(shù)法有 $a_%7Bn%2B1%7D%2B%5Cfrac%7Bp%7D%7Bm-1%7D%3Dm%5Cleft%20(%20%20a_%7Bn%7D%2B%5Cfrac%7Bp%7D%7Bm-1%7D%5Cright%20)%20%2Bk%5Ccdot%20r%5En%20$

形式同7.

9. $a_%7Bn%2B1%7D%3Dma_n%2Bpn%5E2%2Bqn%2Br$

由待定系數(shù)法設(shè)

$a_%7Bn%2B1%7D%2BA%5Cleft%20(%20n%2B1%20%5Cright%20)%5E2%2BB%5Cleft%20(%20n%2B1%20%5Cright%20)%20%2BC%3Dm%5Cleft%20(%20a_%7Bn%2B1%7D%2BA%20n%20%5E2%2BB%20n%20%2BC%20%5Cright%20)%20$