题目内容
计算:
+(
+
)+(
+
+
)+(
+
+
+
)+…+(
+
+
+
+…+
)
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 1 |
| 5 |
| 2 |
| 5 |
| 3 |
| 5 |
| 4 |
| 5 |
| 1 |
| 49 |
| 2 |
| 49 |
| 3 |
| 49 |
| 4 |
| 49 |
| 48 |
| 49 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:本题的关键点在于发现每个括号内分为一组,则:第一组
,第二组
+
=
,第三组
+
+
=
,…第n组的值为
,原算式变形为
+
+
+
+…+
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 2 |
| 2 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 3 |
| 2 |
| n |
| 2 |
| 1 |
| 2 |
| 2 |
| 2 |
| 3 |
| 2 |
| 4 |
| 2 |
| 48 |
| 2 |
解答:
解:
+(
+
)+(
+
+
)+(
+
+
)+…+(
+
+
+
+…+
)
=
+
+
+…+
=
×(1+2+3+…+48)
=
×(49×24)
=588
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 1 |
| 5 |
| 2 |
| 5 |
| 3 |
| 5 |
| 1 |
| 49 |
| 2 |
| 49 |
| 3 |
| 49 |
| 4 |
| 49 |
| 48 |
| 49 |
=
| 1 |
| 2 |
| 2 |
| 2 |
| 3 |
| 2 |
| 48 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=588
点评:此题为数的计算的灵活运用,需要发现各组数的和的特点,针对特点进行运算.
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