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)猜想并写出:
=
-
-
.
(2)直接写出下列各式的计算结果:
①
+
+
+…+
=
;
②
+
+
+…+
=
.
(3)探究并计算:
+
+
+…+
.
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 |
3 |
4 |
(1)猜想并写出:
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
1 |
n |
1 |
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 |
2008×2009 |
2008 |
2009 |
2008 |
2009 |
②
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
n(n+1) |
n |
n+1 |
n |
n+1 |
(3)探究并计算:
1 |
2×4 |
1 |
4×6 |
1 |
6×8 |
1 |
2008×2010 |
分析:(1)归纳总结得到拆项规律,写出即可;
(2)两式分别利用拆项规律变形,计算即可得到结果;
(3)根据得出的规律变形,计算即可得到结果.
(2)两式分别利用拆项规律变形,计算即可得到结果;
(3)根据得出的规律变形,计算即可得到结果.
解答:解:(1)
=
-
;
(2)①原式=1-
+
-
+…+
-
=1-
=
;
②原式=1-
+
-
+…+
-
=1-
=
;
(3)原式=
×(
-
+
-
+…+
-
)=
×
=
.
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 |
2008 |
1 |
2009 |
1 |
2009 |
2008 |
2009 |
②原式=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
1 |
n+1 |
n |
n+1 |
(3)原式=
1 |
2 |
1 |
2 |
1 |
4 |
1 |
4 |
1 |
6 |
1 |
2008 |
1 |
2010 |
1 |
2 |
502 |
1005 |
251 |
1005 |
故答案为:(1)
1 |
n |
1 |
n+1 |
2008 |
2009 |
n |
n+1 |
点评:此题考查了分式的加减法,弄清题中的规律是解本题的关键.
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