题目内容
已知:
=
-
,
=
-
.由以上两式,可以类比得到:
= .
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n(n+1)(n+2) |
| 1 |
| 2n(n+1) |
| 1 |
| 2(n+1)(n+2) |
| 1 |
| n(n+1)(n+2)(n+3) |
考点:类比推理
专题:计算题,推理和证明
分析:利用裂项法,进行类比,即可得出结论.
解答:
解:∵:
=
-
,
=
-
.
∴
利用裂项法可得:
=
-
.
故答案为:
-
.
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n(n+1)(n+2) |
| 1 |
| 2n(n+1) |
| 1 |
| 2(n+1)(n+2) |
∴
| 1 |
| n(n+1)(n+2)(n+3) |
| 1 |
| n(n+1)(n+2)(n+3) |
| 1 |
| 3n(n+1)(n+2) |
| 1 |
| 3(n+1)(n+2)(n+3) |
故答案为:
| 1 |
| 3n(n+1)(n+2) |
| 1 |
| 3(n+1)(n+2)(n+3) |
点评:本题考查类比推理,解题的关键掌握并理解类比推理的定义,并能根据类比的定义鉴别所举的事例是否满足类比推理.
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