题目内容
如果
+
+
+…+
=
,那么n= .
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n(n+1) |
| 2011 |
| 2012 |
分析:由题意可得:
+
+
+…+
=
+
+
+…+
=1-
=
,则可求得n的值.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n(n+1) |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| n(n+1) |
| 1 |
| n+1 |
| 2011 |
| 2012 |
解答:解:∵
+
+
+…+
=
+
+
+…+
=1-
+
-
+
-
+…+
-
=1-
=
=
,
∴n=2011.
故答案为:2011.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n(n+1) |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| n(n+1) |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n+1 |
| n |
| n+1 |
| 2011 |
| 2012 |
∴n=2011.
故答案为:2011.
点评:此题考查了分式的加减运算的应用问题,此题属于规律性题目,难度适中.注意解此题的关键是得到
+
+
+…+
=
+
+
+…+
=1-
+
-
+
-
+…+
-
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n(n+1) |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| n(n+1) |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
练习册系列答案
相关题目