题目内容
有一组数:
,
,
,
,
,
,
,
,
,
,…
,
,…,
,这组数的和是
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 4 |
| 4 |
| 1 |
| 100 |
| 2 |
| 100 |
| 100 |
| 100 |
2575
2575
.分析:由题意义可知,此算式为:
+
+
+
+
+
+
+
+
+
+…+
+
+…+
,
通过试算可以发现:
=1,
+
=1.5,
+
+
=2,
+
+
+
=2.5,…,
+
+…+
=50.5,由此可得:
原式=1+1.5+2+2.5++2+3.5+…50.5=(1+2+3+…+50)+(1.5+2.5+3.5+…+50.5),然后根据等差数列的求和公式进行计算即可.
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 4 |
| 4 |
| 1 |
| 100 |
| 2 |
| 100 |
| 100 |
| 100 |
通过试算可以发现:
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 4 |
| 4 |
| 1 |
| 100 |
| 2 |
| 100 |
| 100 |
| 100 |
原式=1+1.5+2+2.5++2+3.5+…50.5=(1+2+3+…+50)+(1.5+2.5+3.5+…+50.5),然后根据等差数列的求和公式进行计算即可.
解答:解:
+
+
+
+
+
+
+
+
+
+…+
+
+…+
,
=
+(
+
)+(
+
+
)+(
+
+
+
)+…+(
+
+…+
),
=1+1.5+2+2.5++2+3.5+…50.5,
=(1+2+3+…+50)+(1.5+2.5+3.5+…+50.5),
=(1+50)×50÷2+(50.5+1.5)×50÷2,
=1275+1300,
=2575.
故答案为:2575.
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 4 |
| 4 |
| 1 |
| 100 |
| 2 |
| 100 |
| 100 |
| 100 |
=
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| 4 |
| 4 |
| 1 |
| 100 |
| 2 |
| 100 |
| 100 |
| 100 |
=1+1.5+2+2.5++2+3.5+…50.5,
=(1+2+3+…+50)+(1.5+2.5+3.5+…+50.5),
=(1+50)×50÷2+(50.5+1.5)×50÷2,
=1275+1300,
=2575.
故答案为:2575.
点评:通过观察计算,发现式中数据的特点及内在规律,并根据规律选择合适的方法进行计算是完成本题的关键.
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