题目内容
观察下列等式:
=1-
•
=
-
•
=
-
,将以上三个等式两边分别相加,得
+
+
=1-
+
-
+
-
=1-
=
.
(1)猜想并写出:
= .
(2)直接写出下列各式的计算结果:
①
+
+
+…+
= ;
②
+
+
+…+
= .
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 3 |
| 4 |
(1)猜想并写出:
| 1 |
| n(n+1) |
(2)直接写出下列各式的计算结果:
①
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2014×2015 |
②
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| n(n+1) |
考点:有理数的混合运算
专题:计算题
分析:(1)根据题意得出拆项规律,写出即可;
(2)利用拆项法把各式变形,计算即可得到结果.
(2)利用拆项法把各式变形,计算即可得到结果.
解答:解:(1)根据题意得:
=
-
;
(2)①原式=1-
+
-
+…+
-
=1-
=
;
②原式=1-
+
-
+…+
-
=1-
=
.
故答案为:(1)
-
;(2)①
;②
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
(2)①原式=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2014 |
| 1 |
| 2015 |
| 1 |
| 2015 |
| 2014 |
| 2015 |
②原式=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n+1 |
| n |
| n+1 |
故答案为:(1)
| 1 |
| n |
| 1 |
| n+1 |
| 2014 |
| 2015 |
| n |
| n+1 |
点评:此题考查了有理数的混合运算,熟练掌握运算法则是解本题的关键.
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