题目内容
计算:
+
+
+…+
=
.
| 1 |
| 3+6 |
| 1 |
| 3+6+9 |
| 1 |
| 3+6+9+12 |
| 1 |
| 3+6+9+12+…+150 |
| 49 |
| 153 |
| 49 |
| 153 |
分析:此题算式较长,如果按常规方式来解答,是行不通的.通过观察,分母中的数字很有特点,先把
提出来,通过转化,分母变为n(n+1)的形式,然后化为两个分数相减的形式,最后通过分数加、减相互抵消,得出结果.
| 1 |
| 3 |
解答:解:
+
+
+…+
,
=
×(
+
+
…+
),
=
×[
+
+
…+
],
=
×[
-
+
-
+
-
…+
-
],
=
×[
-
],
=
×
,
=
| 1 |
| 3+6 |
| 1 |
| 3+6+9 |
| 1 |
| 3+6+9+12 |
| 1 |
| 3+6+9+12+…+150 |
=
| 1 |
| 3 |
| 1 |
| 1+2 |
| 1 |
| 1+2+3 |
| 1 |
| 1+2+3+4 |
| 1 |
| 1+2+3+4+…+50 |
=
| 1 |
| 3 |
| 2 |
| 2×(2+1) |
| 2 |
| 3×(3+1) |
| 2 |
| 4×(4+1) |
| 2 |
| 50×(50+1) |
=
| 2 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 50 |
| 1 |
| 51 |
=
| 2 |
| 3 |
| 1 |
| 2 |
| 1 |
| 51 |
=
| 2 |
| 3 |
| 49 |
| 102 |
=
| 49 |
| 153 |
点评:此题主要通过提取
,转化为
,得到和的
,化为分数加、减抵消,解出结果.
| 1 |
| 3 |
| 1 |
| n×(n+1) |
| 2 |
| 3 |
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