题目内容
(1)1
+2
+3
+4
+5
;
(2)1
+2
+3
+4
+5
+6
+7
+8
+9
+10
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
(2)1
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
| 1 |
| 512 |
| 1 |
| 1024 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:这两道题属于一种类型,县把每个分数进行拆分,然后分数与整数分别相加,再把每个分数拆成两个分数相减的形式,通过加减相互抵消,求出结果.
解答:
解:(1)1
+2
+3
+4
+5
=(1+2+3+4+5)+(
+
+
+
+
)
=15+(1-
+
-
+
-
+
-
+
-
)
=15+(1-
)
=15+
=15
(2)1
+2
+3
+4
+5
+6
+7
+8
+9
+10
=(1+2+3+…+10)+(
+
+
+…+
)
=(1+10)×10÷2+(1-
+
-
+
-
-
)
=55+(1-
)
=55+
=55
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
=(1+2+3+4+5)+(
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
=15+(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 6 |
=15+(1-
| 1 |
| 6 |
=15+
| 5 |
| 6 |
=15
| 5 |
| 6 |
(2)1
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
| 1 |
| 512 |
| 1 |
| 1024 |
=(1+2+3+…+10)+(
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 1024 |
=(1+10)×10÷2+(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 512 |
| 1 |
| 1024 |
=55+(1-
| 1 |
| 1024 |
=55+
| 1023 |
| 1024 |
=55
| 1023 |
| 1024 |
点评:此题解答的关键在于分数的拆分,然后简算即可.
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