题目内容
记an为(1+x)n+1的展开式中含xn-1项的系数，则
(
+
+…+
)=______．
| lim |
| n→∞ |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
由题意可得 an=
=
=
，
∴
=
=2（
-
），
∴
(
+
+…+
)=
2[（
-
）+（
-
）+（
-
）+…+（
-
）]=
2（1-
）=2，
故答案为：2．
| C | n-1n+1 |
| C | 2n+1 |
| n(n+1) |
| 2 |
∴
| 1 |
| an |
| 2 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴
| lim |
| n→∞ |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| lim |
| n→∞ |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
| lim |
| n→∞ |
| 1 |
| n+1 |
故答案为：2．
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