题目内容
(2013?成都模拟)
+
+
+
+
+
+
+
+…+
+
+
+…+
+
+
+…+
+
.
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 2 |
| 100 |
| 3 |
| 100 |
| 99 |
| 100 |
| 100 |
| 100 |
| 99 |
| 100 |
| 2 |
| 100 |
| 1 |
| 100 |
分析:通过分析发现,
+
+
=2,
+
+
+
+
=3,…,
+
+
+…+
+
+
+…+
+
=100.即算式中分母相同的数相加的和构一个等差数列2,3,…,100,由此可将原式变为等差数列和相加的形式进行巧算
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 2 |
| 100 |
| 3 |
| 100 |
| 99 |
| 100 |
| 100 |
| 100 |
| 99 |
| 100 |
| 2 |
| 100 |
| 1 |
| 100 |
解答:解:
+
+
+
+
+
+
+
+…+
+
+
+…+
+
+
+…+
+
=(
+
+
)+(
+
+
+
+
)+…+(
+
+
+…+
+
+
+…+
+
)
=2+3+4+…+100
=(2+100)×99÷2
=5049.
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 2 |
| 100 |
| 3 |
| 100 |
| 99 |
| 100 |
| 100 |
| 100 |
| 99 |
| 100 |
| 2 |
| 100 |
| 1 |
| 100 |
=(
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 2 |
| 100 |
| 3 |
| 100 |
| 99 |
| 100 |
| 100 |
| 100 |
| 99 |
| 100 |
| 2 |
| 100 |
| 1 |
| 100 |
=2+3+4+…+100
=(2+100)×99÷2
=5049.
点评:本题利用到了等差数列的求和公式:等差数列的和=(首项+尾项)×项数÷2.
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