折紙與錯(cuò)位相減（2021新高考Ⅰ，16）

2023-04-18 13:07 作者:數(shù)學(xué)老頑童 0人讀過 | 我要投稿

（2021新高考Ⅰ，16）某校學(xué)生在研究民間剪紙藝術(shù)時(shí)，發(fā)現(xiàn)剪紙時(shí)經(jīng)常會(huì)沿紙的某條對(duì)稱軸把紙對(duì)折.規(guī)格為 $20%20%5Cmathrm%7Bdm%7D%5Ctimes%2012%20%5Cmathrm%7Bdm%7D$ 的長方形紙，對(duì)折 $1$ 次共可以得到 $10%20%5Cmathrm%7Bdm%7D%5Ctimes%2012%20%5Cmathrm%7Bdm%7D$ 、 $20%20%5Cmathrm%7Bdm%7D%5Ctimes%206%20%5Cmathrm%7Bdm%7D$ 兩種規(guī)格的圖形，它們的面積之和 $S_1%3D240%20%5Cmathrm%7Bdm%7D%5E2$ ，對(duì)折 $2$ 此共可以得到 $5%20%5Cmathrm%7Bdm%7D%5Ctimes%2012%20%5Cmathrm%7Bdm%7D$ 、 $10%20%5Cmathrm%7Bdm%7D%5Ctimes%206%20%5Cmathrm%7Bdm%7D$ 、 $20%20%5Cmathrm%7Bdm%7D%5Ctimes%203%20%5Cmathrm%7Bdm%7D$ 三種規(guī)格的圖形，它們的面積之和 $S_2%3D180%20%5Cmathrm%7Bdm%7D%5E2$ ，以此類推.則對(duì)折 $4$ 次共可以得到不同規(guī)格圖形的種數(shù)為________；如果對(duì)折 $n$ 次，那么 $%5Csum_%7Bk%3D1%7D%5En%7BS_k%7D%3D$ ________ $%5Cmathrm%7Bdm%7D%5E2$ .

解：對(duì)折 $4$ 次，共有

橫 $0$ 縱 $4$ ，橫 $1$ 縱 $3$ ，橫 $2$ 縱 $2$ ，橫 $3$ 縱 $1$ ，橫 $4$ 縱 $0$ ，

$5$ 種不同的折法，可以得到 $5$ 種不同規(guī)格的圖形

易知 $S_k%3D240%5Ccdot%5Cleft(%20k%2B1%20%5Cright)%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%5Ek%7D$ ，所以

$%5Cbegin%7Baligned%7D%0A%09%5Csum_%7Bk%3D1%7D%5En%7BS_k%7D%26%3D%5Csum_%7Bk%3D1%7D%5En%7B%5Cleft%5B%20240%5Cleft(%20k%2B1%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Ek%20%5Cright%5D%7D%5C%5C%0A%09%26%3D240%5Csum_%7Bk%3D1%7D%5En%7B%5Cleft%5B%20%5Cleft(%20k%2B1%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Ek%20%5Cright%5D%7D%5C%5C%0A%09%26%3D240%5Cleft%5B%202%5Ctimes%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E1%2B3%5Ctimes%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B%5Ccdots%20%2B%5Cleft(%20n%2B1%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5En%20%5Cright%5D%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bk%3D1%7D%5En%7BS_k%7D%3D240%5Cleft%5B%202%5Ctimes%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B3%5Ctimes%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E3%2B%5Ccdots%20%2B%5Cleft(%20n%2B1%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%20%5Cright%5D%20$

兩式相減，可得

$%5Cbegin%7Baligned%7D%0A%09%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bk%3D1%7D%5En%7BS_k%7D%26%3D240%5Cleft%5B%201-%5Cleft(%20n%2B1%20%5Cright)%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%2B%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B%5Ccdots%20%2B%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5En%20%5Cright%5D%5C%5C%0A%09%26%3D240%5Cleft%5B%201-%5Cleft(%20n%2B1%20%5Cright)%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%2B%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D-%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cright%5D%5C%5C%0A%09%26%3D240%5Cleft%5B%201-%5Cleft(%20n%2B1%20%5Cright)%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7B2%7D-2%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E%7Bn%2B1%7D%20%5Cright%5D%5C%5C%0A%09%26%3D240%5Cleft(%20%5Cfrac%7B3%7D%7B2%7D-%5Cfrac%7Bn%2B3%7D%7B2%5E%7Bn%2B1%7D%7D%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$