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# (教案)递推数列求通项公式

33 99 3 2 5 3 7 4 (2) 7 5 13 8 19 11 (3)1 ,0 ,-1 ,0 ,1 ,0 ,-1,0 (4)1,3,7,15,31

1 3

9 35

17 63

3 2

1 5 n2 n a = , b = 0, c = , an = + 1 6 6 2 2

n n n

n

n

n +1

= 2( S n 3n )
n 1

1 n(n + 1) 且Sn = an 数列 {an } 的通项公式? 6 2 10 n ) ( n ∈ N * )试求该数列 {an } 有 11

n 98 (n ∈ N * ) ,则该数列在前 30 项中,最大项与最小项是 n 99

4n 3 4n 2

1 1 , an +1 an = 求数列 {an } 的通项公式? 2 2 4n 1

2. an +1 = f ( n) an 法②

an =

an an 1 a 2 a1 = f ( n 1) f ( n 2) f (1) a1 an 1 an 2 a1

λ = 1 an +1 + 1 = 2(an + λ )

n 1

= 2 2 n 1 an a n = 2 a n 1 + 1 … ②,故① ②

n

n

an 1 an 1 an 1 + n 1 = 1 ,令 bn = n b n + bn 1 = 1 转化为3类型. n 3 3 3 3 3
n

2

an +1 = 2 S n 2n n 2 n 1 2 an = 2 S n 1 2 (n 1) , (n ≥ 2)
n 1

2n + 1
n +1

q (n + 1) + b = 3(an + p 2 n qn + b) 则

1 , q = 1, b = 0 2

{a

n

a, (n = 1) 2n n}以3为公比, 则 an = n2 n 1 (2a 7)3 + 2 + n, (n ≥ 2)
an +1 an 2 n 1 2n 1 an +1 2n 1 2n 1 = n n +1 n +1 ,设 bn +1 = n +1 ,由 bn +1 bn = n +1 n +1 3n +1 3 3 3 3 3 3
(注右边当作两数列,等比,与等比差数列,故能求和)

5.分式递推数列,一般取"倒"的方法: 形式 an +1 =

can ban + d

3an 1 , an +1 = , 求数列 {an } 的通项公式? 2 2 an + 1
n +1

2a + 1 2 1 1 1 1 = n = + ,令 bn = 则有 b an +1 3an 3 3 an an

1 2 = bn + 转化为3类型 . 3 3

6.( 第 5 类 型 变 形 ) a n +1 a n = pa n +1 qa n 类 型 , 一 般 处 理 为 : 若 p = q , 则 转 化 为

1 1 1 1 p 1 1 = 从而为等差数列 .若 p ≠ q ,则可化为 = ,即转化为类型 3. an +1 an p an +1 q an q

1 , 2

a n a n +1 = 2 a n a n +1 ,求数列 {an } 的通项公式? 1 =

1 1 1 = ( + 1) , an +1 2 an

∴a1

n +1

1 1 ( 1) , 2 an

∴ a1 1 是首项为 a1 1 = 1 ,公比为 1 的等比数列. 2

n

1

∴a1 1 = 21
n

n 1

2n 1 2n 1 + 1
1 2

2

{an } 的通项公式?

2

∵ S = a = 1 ≠ 0 ,∴S
1 1

1 2

n

≠0

∴S1 S1
n

= 2,

n 1

∴ S1 是以 2 为公差, S1 = 1为首项的等差数列.

n

1

∴ S1 = 2n 1 , ∴S
n

n

=

1 2n 1

1 1 2 = 2n 1 2n 3 (2n 1)(2n 3)

1(n = 1) 故 an = 2 (2n 1)(2n 3) (n ≥ 2)
7(了解). an +1 = 形式. 例 7 .已知数列 {an } 满足 a1 =

f (n)an 类型,一般为等式两边取倒数后转化为 an +1 = pan + q 的 g (n)an + d (n) 3(n + 1)an 3 , an +1 = ,求数列 {an } 的通项公式? 2 2 an + n

:

1 n 2 = + an +1 3(n + 1)an 3(n + 1)

.

n +1 1 n 2 n +1 1 n = + .设 + λ = ( + λ ) 可 求 得 λ = 1 , a n +1 3 an 3 a n +1 3 an

∴ an 1

n

1 1 1 1 = 为 首 项 , 为 公 比 的 等 比 数 列 . a1 3 3

n 1 1 1 = ( ) n 1 , 整 理 得 an 3 3

n 3n an = n 3 1
8(了 解 ). 对于 an +1 =

aan + b 类 型 , 一 般采 用 待定 系数 法 , 转化 为等 式 两边 取倒 数 , 变为 can + d ax + b 的根求系数. cx + d

bn +1 = pbn + q 的形式. 也可用特征方程 x =

1 an 1 2

,

1 1 + ,代入递推关系 8 a n + 1 a n 1 6 a n + 1 + 2 a n + 5 = 0 ,整 bn 2

4 6 3 4 + = 0 即 bn +1 = 2bn , bn +1bn bn +1 bn 3

∴b
n

n +1

4 4 = 2(bn ) . 3 3
n 1

∴ b 4 是首项为,公比为 2 的等比数列.故 b 4 = 2 2 3 3 3

n

,即 bn =

1 n (2 + 4) . 3

an =

1 1 + , bn 2

∴a b = 1 b + 1 = 1 2 2 3
n n n

n 1

5 + . 3

S n = a1b1 + a2b2 + + an bn 1 (1 2n ) 5 1 3 = + n = (2n + 5n 1) 1 2 3 3

1 1 an 2

2 an + 5 2x + 5 1 5 ,故 x = 解有 x1 = , x2 = 16 8an 16 8 x 2 4 5 12(an ) 5 4 ①, an +1 = 4 16 8an

1 6(an ) 1 2 即 an +1 = 2 16 8an an +1

1 1 1 1 a a1 an 2 1 n 2 2= 2 = 2 为首项, 以 则①②相除,有 ,故数列 是以 5 2 5 5 5 an an +1 an a1 4 4 4 4
1 为公比的数列, 2

2n 1 + 5 . 2n + 4

1 3 1 1 = n ,则 bn = = (2n + 4) 1 3 2 2 +4 an 2

an +1 =

2 an + 5 16 8an

, 设

an +1 + λ =

2 an + 5 +λ 16 8an

, 得

an +1 + λ =

(2 8λ )an + 5 + 16λ (2 8λ )(an + λ ) + 5 + 14λ + 8λ 2 , an +1 + λ = 16 8an 16 8an

2

1 5 ,λ = 2 4

1 5 6(an ) 12(an ) 1 1 5 5 2 ①,当 λ = 时,有 a = 4 当 λ = 时,有 an +1 = n +1 2 2 16 8an 4 4 16 8an an +1

1 1 1 1 an a1 an 2 2 = 2 为首项, 以 1 2=1 2 ,故数列 则①②相除,有 是以 5 2 5 5 5 2 an an +1 an a1 4 4 4 4

2n 1 + 5 2n + 4
2 ,求数列 {an } 的通项公式? an + 1

λ ( an + λ ) λ 2 + λ + 2 2 解,( 带待定系数法) an +1 + λ = + λ an +1 + λ = an + 1 an + 1

2

2(an + 2) 1 1 1 1 ,化简向类型 5 转化. = + an + 1 an +1 + 2 2 (an + 2) 2

1 1 1 (2) n +1 + 2 令 bn = 向类型 3 转化: bn +1 = bn + .再求解. an = 2 2 (2) n +1 1 an + 2
9

a n +1 = pa n q

( p , an > 0 )类型,常用对数转化.

lg an+1 = lg p + q lg an令bn = lg an , 得 bn+1 = qbn + lg p 转 化

an+2 an+1

an+2 = a n +1

a n +1 , 求数列 {an } 的通项公式? an

lg

a 1 a = lg n+1 lg n+1 是以 q = 1 2 an 2 an
1

a2 = lg a1

3

1 ( ) n 1 3( ) n 1 a n +1 3 2 2 = lg ∴ lg a = lg n

0 1 2 1n an+1 21n a a a = 3 an = a1 2 3 n = 1 32 32 32 32 an a1 a2 an1

= 32 +2
0

1

+22 ++21n

=3

1 1 2 1 1 2

n1

.

10. (二次一阶递推数列)一般分解因式降次,为能否化生为熟. 例 11. ①在数列 {an } 中, a1 = 1 , an > 0 , an +1 an an +1 = 2an .求数列 {an } 的通项公式?
2 2

②在数列 {an } 中, a1 = 1 , an > 0 , ( n + 1) an +1 + an an +1 = nan .求数列 {an } 的通项公式?
2 2

a1 = 1, a2 = 6 ( p , q 为常数).常向等比数列转化,用带待定系数. an + 2 + pan +1 + qan = 0

an +1 + λ an = k (an + λ an 1 ) an +1 = (k λ )an + k λ an 1

,

k λ = 5 λ = 2 k λ = 6 k = 3

λ = 3 an +1 2an = 3(an 2an 1 ) k = 2

an +1 3an = 2(an 3an 1 ) 均为等比数列.

∴a

n +1

2an = 3 3n 1 = 3n ,

∴a

n +1

3an = 2 2n 1 = 2n ,两式相减: an = 3n 2n .

λ = 2 得 an +1 2an = 3( an 2an 1 ) ,这是以 a2 a1 为首项 3, q = 3 的等 k = 3
n +1

∴a

2an = 3 3n 1 = 3n an +1 = 2an + 3n , (用迭代法)
( an = 2
n2

(3n + 1) )

11(了解周期数列). an +T = an ,(T ≠ 0 ).对任意的正整数都成立,可利用周期性解决. 例 13.. 在数列 {an } 中, a1 = 13 , a2 = 56 ,对所以的正整数 n 有: an +1 = an + an + 2 求 a2009 ? 解,由 an +1 = an + an + 2 故 an + 2 = an +1 + an + 3 ,两式相加. an + 3 = an ,

∴a

n+ 6

= an + 3 = an

∴ {a } 是以 6 为周期. ∴ a
n

2009

= a334×6+5 = a5 ,经计算 a2009 = 56 .

12.归纳猜想型 例 14:已知数列 {an } 满足 an +1 = an nan + 1, a1 = 2, 求 a2 , a3 , a4 猜想 an 并证明你的结论.
2

2 2 2

2 2

1 1 mn 1 f ( ) f ( ) = f ( ) . 记 an = f ( 2 ) n∈ N , 则 数 列 m n 1 mn n + 5n + 5

{an }

1 f( ) 2

B f( )

1 3

C f( )

1 4

D f( )

1 5

1 (n + 2) (n + 3) 1 1 )= f( )= f( ) f ( ) ,再由迭加法.2 1 (n + 2)(n + 3) n + 5n + 5 n+2 n+3
2

2. . 已 知 函 数 f ( x ) =

3x , ( x ∈ R ) , 正 项 等 比 数 列 {an } 满 足 a50 = 1 , 则 3x + 1
C

f (ln a1 ) + f (ln a2 ) + f (ln a99 ) =
A 99 分 B 101 析 C :

99 2

D 因

101 2

f ( x) + f ( x) = 1

,

f (ln a99 ) = f (ln

1 ) = f ( ln a1 )即f (ln a99 ) + f (ln a1 ) = 1 . a1
2010 5an 1 2 (n ≥ 2, n ∈ N ) ,数列的前 2010 项的和为 403,则 ∑ ai ai +1 an 1 5 i =1

3. 数列 {an } 满足 an =

2010 i =1

5a 2 .求得 a3 = a 所以数列周期为 2 a5

5an 1 2 an 1an = 5(an 1 + an ) 2 an 1 5 = 5 ∑ (ai + ai +1 ) 2 × 2010 = 5 × 2 ∑ ai 2 × 2010 = 10
i =1 i =1 2010 2010

∑aa

i i +1

4.已知两个等差数列 {an } , {bn } 的前 n 项和 Sn , Tn ,且

S n 3n + 2 a 41 = ,则 7 = Tn 2n + 1 b5 19 S n 2n + 3 a 37 = ,则 9 = Tn 3n 1 b9 50

5.已知两个等差数列 {an } , {bn } 的前 n 项和 Sn , Tn ,且 注意:3,4 的解法是有区别的

6.已知数列 {an } 的前 n 项和 Sn 满足, 3S n 4an = 2n 4, n ∈ N , (1) 证明:当 n ≥ 2 时, an = 4an 1 2 (2) 求数列的 {an } 的通项公式. (3) 设 cn =

an 2n + 1 , Tn 为数列 {cn } 的前 n 项和,证明: Tn < an +1 8
①,得当 n ≥ 2 时, 3S n 1 4an 1 = 2( n 1) 4, ②.

2 3

2 4 = . 3 3

4 n 1 4 n 4n + 2 4 = an = ,n∈ N 3 3 3

(3) cn =

an 4n + 2 4n + 2 1 2 = n +1 < n +1 = + n +1 an +1 4 + 2 4 4 4 a1 1 3 = < a2 3 8

a1 n 1 1 1 1 + + 2( 3 + 4 + + n +1 ) a2 4 4 4 4

1 1 1 n +1 i 3 1 n 1 = + + 2 4 4 4 1 3 4 1 4 2n + 1 2 2n + 1 = < n +1 8 3i4 8

7. 已知数列 {an } 的前 n 项和 Sn , m = ( S n , n + 1) , n = ( ,

1 1 ) ,且 an +1 = m n , a1 = 1 .求 n 2n

an 的通项公式.

Sn n + 1 S n +1 + n S n +1 S n = n + n n 2 n 2 解: S S 1 n +1 n = n n +1 n 2 an +1 = S S S S S 2 S1 S 2 S1 = + ( ) + ( 3 2 ) + + ( n n 1 ) 2 1 2 1 3 2 n n 1 所以: 1 1 1 1 = 1 + + 2 + + n 1 = 2 1 ( ) n 2 2 2 2 Sn n + 1 n 1 + n = 2+ n n 2 2 n2 当 n ≥ 2 时, an = 2 + n 1 ,又当 n = 1 时, a1 = 1 2 n2 所以 an = 2 + n 1 2 2 8.已知数列 {an } 的前 n 项和 S n ( n ∈ N ) , a1 = ,且当 n ≥ 2 时. S n S n 1 3S n + 2 = 0 3

1 ,求数列 {bn } 的通项公式 Sn 1

(3)设数列

Sn +1 1 n 1 n 的前 n 项和 Tn ,证明 < Tn < 2 7 2 Sn 1

2 4 8 ,易知 a2 = , a3 = 3 21 105 2 6 14 30 62 , S 4 = , S5 = (2) 由(1)知: S1 = , S 2 = , S3 = 3 7 15 31 63
(1) 由 a1 =

2n +1 2 2n +1 1
2 k +1 2 ,当 n = k + 1 时,由 S k +1S k 3S k +1 + 2 = 0 2k +1 1

2 2( k +1)+1 2 = ( k +1) +1 即证. 3 Sk 2 1

1 1 ,故 bn = = 2n +1 + 1 Sn 1 2 1
n +1

1 S n +1 1 2n +1 1 1 (3) 令 cn = = n+2 = n +2 Sn 1 2 1 2 2 2 1 1 n 则 Tn = c1 + c2 + c3 + cn > 即证 2 2 1 1 1 1 1 7 1 1 8 1 1 1 又 cn = = > = n+ 2 n+2 n+2 2 2 2 1 2 14 2 1 2 14 2 2 7 2n

n 1 1 1 2 1 3 1 + ( ) + ( ) + + ( )n 2 7 2 2 2 2 1 1 n 1 ( 2 ) n 1 n 1 2 > 即证. = 1 2 7 2 7 1 2

2 3 S n 1

S 1 2 1 = n 1 3 S n 1 3 S n 1

3 S n 1 1 2 = = 1 + S n 1 S n 1 1 S n 1 1

1 Sn 1

n +1

+1

1 1 得 Sn = 1 + 代 入 S n S n 1 3S n + 2 = 0 中 , 化 间 有 Sn 1 bn
(点拨,以后如此递推关系可以考虑此法)

bn = 2bn 1 1 再解之.

S n +1 1 2 n +1 1 1 1 1 1 = n+2 = = (3)由 n n S n 1 2 1 2 8i2 2 2 7i2 + 2n 2

S 1 1 1 1 n 1 1 n ≤ n +1 < 从而 (1 n ) ≤ Tn < n 2 7i 2 Sn 1 2 2 7 2 2

n 1 n < Tn < 即证. 2 7 2
9.在数列 {an } , {bn } 中, a1 = 2, b1 = 4 且 an , bn , an +1 成等差数列, bn , an +1 , bn +1 成等比数列. (1) 求 a2 , a3 , a4 及 b2 , b3 , b4 ,由此猜想 {an } , {bn } 的通项公式,并证明 (2) 证明:

1 1 1 5 + + + < a1 + b1 a2 + b2 an + bn 12
2

2

2 2

bk +1 =

ak2+1 = (k + 2) 2 ,所以 n = k + 1 时,也成立. bk
2

1 1 5 = < , a1 + b1 6 12

1 1 1 1 1 1 1 1 + + + < + ( + + + ) a1 + b1 a2 + b2 an + bn 6 2 2 × 3 3 × 4 n × (n + 1)

1 1 1 1 1 1 1 1 + ( + + + ) 6 2 2 3 3 4 n n +1 1 1 1 1 1 1 5 = + ( )< + = 即证. 6 2 2 n + 1 6 4 12 =

### 高中数学必修五北师大版 数列的通项公式与递推公式 教案

2.1.2 数列通项公式递推公式 一、教学目标: 1.了解数列递推公式,明确递推公式与通项公式的异同; 2.会根据数列递推公式写出数列的前几项; 3.理解...