题目内容
(2012?长清区模拟)1
+2
+3
+4
+…+20
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 420 |
分析:本题由各加数的特征可以看出整数部分是一个公差为1的等差数列,分数部分分母可以表示为:1×2,2×3,3×4,4×5…20×21.那么原式1
+2
+3
+4
+…+20
=1
+2
+3
+4
+…+20
.那么通过式的变形及分数拆项求解即可.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 420 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 20×21 |
解答:解:1
+2
+3
+4
+…+20
=1
+2
+3
+4
+…+20
.
=(1+2+3+4+…+20)+(1-
+
-
+
-
+…+
-
)
=(1+20)×10+1-
=210+
=210
.
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 420 |
=1
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 20×21 |
=(1+2+3+4+…+20)+(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 20 |
| 1 |
| 21 |
=(1+20)×10+1-
| 1 |
| 21 |
=210+
| 20 |
| 21 |
=210
| 20 |
| 21 |
点评:此题是一个难度较高的等差数列与可拆项分数复合求和的题目.难点在于分数的巧妙分拆.多加练习此类题目就能驾熟就轻了奥!
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