题目内容
|
…
观察这组等式的规律,完成下列各题
(1)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 10×11 |
(2)若
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 11×13 |
| n |
| 13 |
分析:(1)利用
=
-
计算即可;
(2)利用
=
×(
-
)先化简,再解关于n的方程求解即可;
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
(2)利用
| 1 |
| n(n+2) |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+1 |
解答:解:(1)
+
+…+
,
=1-
+
-
+…+
-
,
=1-
,
=
;
(2)
+
+…+
+
=-1,
×( 1-
)+
×(
-
)+…+
×(
-
)+
=-1,
-
×
+
=-1,
13-1+2n=-26,
n=-19.
故n的值为19.
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 10×11 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 10 |
| 1 |
| 11 |
=1-
| 1 |
| 11 |
=
| 10 |
| 11 |
(2)
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 11×13 |
| n |
| 13 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 11 |
| 1 |
| 13 |
| n |
| 13 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 13 |
| n |
| 13 |
13-1+2n=-26,
n=-19.
故n的值为19.
点评:本题考查了分数的拆分运算,解题关键是将一个分数拆分成两项,再两两抵消,达到化简的目的.
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