# 求递推数列通项公式的常用方法

an = S n ? S n ?1 (n ≥ 2) ，等差数列或等比数列的通项公式。

*

1 1 an ，又 a1 = ， 2 2

?1? ∴ an = ? ? . ?2?

( Sn +1)

n

= n (n = 1, 2 ???) .试证数列

{an } 是等比数列.

n

-1-

?1? ? ， 若 数 列 {bn } 满 足 b1 = 1 ， ?2?

n

(n ≥ 1) ，求数列 {bn } 的通项公式.

1 ?1? 【解析】 b1 = 1 , bn +1 ? bn = ? ? (n ≥ 1) ,∴ bn = b1 + (b2 ? b1 ) + ???(bn ? bn ?1 ) =1+ + ?? + ： 2 ?2?

n

?1? ? ? ?2?

n?1

?1? =2?? ? ?2?

n?1

.

1 ?1? * 跟踪训练 3.已知 a1 = , an +1 = an + ? ? ( n ∈ N ) ,求数列 {a n } 通项公式. 2 ?2?

n

a a2 a3 ??? n (an ≠ 0, n ≥ 2) 求通项公式的方法称为累乘法, a1 a2 an ?1

an +1 n + 1 a a a = ,又有 an = a1 2 3 ??? n ( an ≠ 0, n ≥ 2) = an n a1 a2 an ?1

2 3 1 2

n = n ,当 n = 1 时 a1 = 1 ，满足 an = n ，∴ an = n . n-1

{a n } 的通项公式是.

(an +1 + x) = q (an + x) 与原递推公式恒等变成 an +1 +

d d = q ( an + ) 的方法叫构造新 q ?1 q ?1

-2-

a1 + 1 = 2 ,公比为 2 的等比数列,即 an + 1 = 2 n ,∴ an = 2n ? 1

n ?1

+ an-1 (n ≥ 2) 求数列 {an } 的通项公式.

can 1 d 1 1 (c ≠ 0, d ≠ 0) ,取倒数变成 = + 的形 an + d an +1 c an c

an ,求数列 {an } 的通项公式. 2 an + 1

【解析】:将 an +1 =

?1? an 1 1 1 1 取倒数得: = 2+ ,∵ ? = 2 , ∴ ? ? 是以 2 an + 1 an +1 an an +1 an ? an ?

1 1 1 = 1 为首项,公差为 2 的等差数列. = 1 + 2(n ? 1) ,∴ an = . a1 2n ? 1 an

2 an ,求数列 {an } 的通项公式. an + 2

n
n ?1

， n = 1 时，

a1 = S1 = 9 满足上式. ∴ {an } 的通项公式 an = 9 (10 )

n ?1

,n ≥ 2时

an = 10 为常数,所以 an ?1

{an } 为等比数列.
2. 解:由已知可求 a1 = 1 , a2 = 3 , a3 = 5 ,猜测 an = 2n ? 1 .(用数学归纳法证明).

1 ?1? ?1? 3. 由已知 an +1 ? an = ? ? ,∴ an = a1 + ( a2 ? a1 ) + ( a3 ? a2 ) + ???( an ? an ?1 ) = + ? ? 2 ?2? ?2? ?1? + ??? + ? ? ?2?
n?1

n

2

=

3 ?1? ?? ? 2 ?2?

n?1

.

4. n ≥ 2 时, an = a1 + 2a2 + 3a3 + ??? + ( n ? 1) an ?1 , an +1 = a1 + 2a2 + ??? + ( n ? 1) an ?1 + nan
-3-

an +1 a a a = n + 1 ,∴ 3 = 3 , 4 = 4 , ??? , n = n a3 an a2 an ?1

?1 n = 1 an n! ? ∴ = 3 × 4 × 5 ???×n , a2 = 1 ,∴ an = (n ≥ 2) ,∴ an = ? n ! . a2 2 ?2 n≥2 ?
5. an =

3n ? 1 2

6. an =

2 n +1

-4-

...全归纳:由数列的递推公式求通项公式的常用方法(含答....doc