题目内容
设an是(3-
)n（n=2，3，4，5，…）展开式中x一次项系数，则
(
+
+
+…+
)=______．
| x |
| lim |
| n→∞ |
| 32 |
| a2 |
| 33 |
| a3 |
| 34 |
| a4 |
| 3n |
| an |
展开式的通项为 Tr+1=(-1)r3n-r
x
令
=1得r=2
∴an=3n-2Cn2．
=
=9×
=
=18×(
-
)，
∴
(
+
+
+…+
)
=
{18×[(1-
) +(
-
)+(
-
)+…+(
-
)]}
=
[18×(1-
)]
=18．
故答案为：18．
| C | rn |
| r |
| 2 |
令
| r |
| 2 |
∴an=3n-2Cn2．
| 3n |
| an |
| 3n | ||
|
| 2 |
| n(n-1) |
| 18 |
| n(n-1) |
| 1 |
| n-1 |
| 1 |
| n |
∴
| lim |
| n→∞ |
| 32 |
| a2 |
| 33 |
| a3 |
| 34 |
| a4 |
| 3n |
| an |
=
| lim |
| n→∞ |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n-1 |
| 1 |
| n |
=
| lim |
| n→∞ |
| 1 |
| n |
=18．
故答案为：18．
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