题目内容
(2012?佛山模拟)
,
,
,
…前30个数的和为
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 30 |
| 31 |
| 30 |
| 31 |
分析:首先应确定出前30个数的末项是
,也就是计算
+
+
+
+…+
的和,根据数字特点,每个分数都能拆成两个分数相减的形式,然后通过加、减相互抵消,得出结果.
| 1 |
| 930 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 930 |
解答:解:
,
,
,
…前30个数的末项是
,
+
+
+
+…+
,
=
+
+
+
+…+
,
=(1-
)+(
-
)+(
-
)+(
-
),
=1-
,
=
.
故答案为:
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 930 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 930 |
=
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 30×31 |
=(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 30 |
| 1 |
| 31 |
=1-
| 1 |
| 31 |
=
| 30 |
| 31 |
故答案为:
| 30 |
| 31 |
点评:因为分母是两个连续自然数的乘积,所以可以分解成两个分数相减的形式,通过抵消,即可得出结果.
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