一道中考填空题的多种解法

一道中考填空题的多种解法

河北省顺平县梁洁华希望中学 张海琴

邮编 072250

题目 如图1，用围棋子按下面的规律摆图形，则摆第n 个图形需要围棋子的枚数是 ．（湖北荆州2010年中考试题）

解法一（数字规律法）

我们把每个图形中围棋子的个数数出来：5，8，11，…，由此我们发现规律：2135+?=，2238+?=，23311+?=，…．

故摆第n 个图形需要围棋子的枚数是：23+n ．

解法二（左右弧线分层法）

将第n 个图形用弧线自左至右分割成)1(+n 层，如图2所示：

每层围棋子的个数分别为2，3，3， (3)

故摆第n 个图形需要围棋子的枚数是：n 32+．

解法三（左右直线分层法）

将第n 个图形自左至右分层，如图3所示：

每层有2个围棋子的有)1(+n 层，每层有1个围棋子的有n 层， 第1个 …

第2个

第3个 图

1 第n 个

图2

第n 个

图

3

故摆第n 个图形需要围棋子的枚数是：23)1(2+=++n n n ． 解法四（左右斜线分层法）

将第n 个图形用斜线自左至右分割成)2(+n 层，如图4所示：

每层围棋子的个数分别为1，3，3，…，3，1．

故摆第n 个图形需要围棋子的枚数是：23+n ．

解法五（上下直线分层法）

将第n 个图形用直线自上至下分成3层，如图5所示：

每层围棋子的个数分别为)1(+n ，n ，)1(+n ．

故摆第n 个图形需要围棋子的枚数是：

3)1()1(+=++++n n n n ．

解法六（平行四边形法）

将第n 个图形看成由一个平行四边形和2个围棋子组成的图形，如图6所示：

由于组成平行四边形的围棋子个数为n 3，故摆第n 个图形需要围棋子的枚数是：23+n ．

解法七（梯形法）

…

第n 个

图

4 第n 个 图

5

第n 个 图

6

将第n 个图形看成由一个梯形和2个围棋子组成的图形，如图7所示：

由于组成梯形的围棋子个数为[]n n n 33)1()1(21

=?++-?，

故摆第n 个图形需要围棋子的枚数是：23+n ．

解法八（三角形法）

将第n 个图形看成由n 个三角形和2个围棋子组成的图形，如图8所示：

故摆第n 个图形需要围棋子的枚数是：23+n ．

感悟：小小棋子摆出了一道培养发散思维能力、动手能力、归纳猜想能力、逻辑推理能力的好题，摆出了一道体现“数形结合”思想方法的好题，值得探究和欣赏！

邮箱：spzhqsh@https://www.wendangku.net/doc/5b18080547.html,

第n 个 图

7 第n 个 图

8

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