题目内容
已知数列a0,a1,a2,…,an,…满足关系式(3-an+1)(6+an)=18,且a0=3,则
的值是
(2n+2-n-3)
(2n+2-n-3).
| n |
| i=0 |
| 1 |
| ai |
| 1 |
| 3 |
| 1 |
| 3 |
分析:由已知可得
=
+
,令bn=
+
,则可构造等比数列{bn},可求bn,进而可求
,利用等比数列的求和公式可求
| 1 |
| an+1 |
| 2 |
| an |
| 1 |
| 3 |
| 1 |
| an |
| 1 |
| 3 |
| 1 |
| an |
解答:解:
=
+
,
令bn=
+
,得b0=
,bn=2bn-1,
∴bn=
×2n.
即
=
,
∴
=
(22+23+…+2n+1)-
×n
=
×
-
=
(2n+2-n-4)
| 1 |
| an+1 |
| 2 |
| an |
| 1 |
| 3 |
令bn=
| 1 |
| an |
| 1 |
| 3 |
| 2 |
| 3 |
∴bn=
| 2 |
| 3 |
即
| 1 |
| an |
| 2n+1-1 |
| 3 |
∴
| n |
| i-0 |
| 1 |
| ai |
| 1 |
| 3 |
| 1 |
| 3 |
=
| 1 |
| 3 |
| 4(1-2n) |
| 1-2 |
| n |
| 3 |
=
| 1 |
| 3 |
点评:本题主要考查了利用数列的递推公式构造等比数列求解数列的通项公式,及等比数列的求和公式的应用
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