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如何用數(shù)學(xué)歸納法解決2023北京高考數(shù)學(xué)壓軸?

2023-06-30 00:18 作者:UTF-0_  | 我要投稿

第三問的等價(jià)形式:

當(dāng)?m%3D1?時(shí),顯然成立

假設(shè)當(dāng)?m%3Dk-1?時(shí)成立

則當(dāng)?m%3Dk?時(shí):

若?a_%7Bn%7D%2C%20b_%7Bn%7D?中都有?k?,則只取兩個(gè)?k?即可

若?b_%7Bn%7D?中沒有?k? ,設(shè)?a_%7Bn%7D?中?k?的個(gè)為?f?

當(dāng)?f%3D0?時(shí),由于?b_%7Bn%7D?中也沒有?k?,于是取子數(shù)列可退化到?k-1?時(shí)的情形,顯然成立

假設(shè)當(dāng)?f%3Dw-1?時(shí)成立,此時(shí)

%5Cleft%5C%7B%20a_%7Bn%7D%20%5Cright%5C%7D%20%3Da_%7B1%7D%2Ca_%7B2%7D%2C...%2Ca_%7Bp%7D%2C...%2Ca_%7Bq%7D%2C...%2Ca_%7Bk%7D

%5Cleft%5C%7B%20b_%7Bn%7D%20%5Cright%5C%7D%20%3Db_%7B1%7D%2Cb_%7B2%7D%2C...%2Cb_%7Bs%7D%2C...%2Cb_%7Bt%7D%2C...%2Cb_%7Bk%7D

則當(dāng)?f%3Dw?時(shí):

考慮將?%5Cleft%5C%7B%20a_%7Bn%7D%20%5Cright%5C%7D?中的某一項(xiàng)?a_%7Br%7D?換為?k?

若?r%5Cnotin%20%5Bp%2Cq%5D?,則顯然成立

若?r%5Cin%20%5Bp%2Cq%5D?,則將?%5Cleft%5C%7B%20a_%7Bn%7D%20%5Cright%5C%7D%20%3Da_%7B1%7D%2Ca_%7B2%7D%2C...%2Ca_%7Bp%7D%2C...%2Ca_%7Br%7D...%2Ca_%7Bq%7D%2C...%2Ca_%7Bk%7D

中的?a_%7Br%7D?換為?k?后,“?a 和”?%5Csum_%7Bi%3Dp%7D%5Eqa_%7Bi%7D%20 增大了?k-a_%7Br%7D%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck-1%20%5Cright%5C%7D%20?

若?%5Cleft%5C%7B%20b_%7Bn%7D%20%5Cright%5C%7D?不取“?b?所有和”?%5Csum_%7Bi%3D1%7D%5Ekb_%7Bi%7D%20?, 則再向前或向后取一項(xiàng),此時(shí)“?b?和”增大了?u%2Cu%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck-1%20%5Cright%5C%7D%20?

取?a_%7Br%7D%3Dk-u%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck-1%20%5Cright%5C%7D%20?,成立

若?%5Cleft%5C%7B%20b_%7Bn%7D%20%5Cright%5C%7D?取“?b?所有和”,因?yàn)?a_%7Br%7D%5Cleq%20m-1%3Cm%5Cleq%20%5Csum_%7Bi%3D1%7D%5Ekb_%7Bi%7D%20?,所以“?a?和”至少是兩項(xiàng)的和,

則去掉“?a?和”的首項(xiàng)或末項(xiàng)?v%2Cv%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck-1%20%5Cright%5C%7D%20?

取?a_%7Br%7D%3Dk-v%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck-1%20%5Cright%5C%7D%20?,成立

即當(dāng)?f%3Dw?時(shí)成立

由數(shù)學(xué)歸納法可知,此時(shí)對(duì)任意?f%5Cin%20%5Cleft%5C%7B%201%2C2%2C...%2Ck%20%5Cright%5C%7D%20?成立

于是當(dāng)?m%3Dk?時(shí)成立

再由數(shù)學(xué)歸納法,于是原命題對(duì)任意?m?成立


如何用數(shù)學(xué)歸納法解決2023北京高考數(shù)學(xué)壓軸?的評(píng)論 (共 條)

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