2023浙江大學(xué)強(qiáng)基數(shù)學(xué)逐題解析（7）

2023-07-04 12:58 作者:CHN_ZCY 0人讀過 | 我要投稿

封面：冬の白ボブの魔法使い。

作畫：ノーコピーライトガール

https://www.pixiv.net/artworks/95091296

18. 已知 $C%2CL%5Cin%5Cmathbb%7BR%7D$ 且 $L%20%5Cneq%200$ ，有

$%5Clim_%7Bn%5Cto%5Cinfty%7D%20%7B%5Cfrac%7Bn%5EC%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Csin%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%7D%7B%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Ccos%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%7D%7D%3DL$

則 $L%3D$ ___________.

答案?? $%5Cfrac%7B2%7D%7B%5Cpi%7D$

解析??

設(shè)

$%5Cbegin%7Baligned%7D%0Aa_n%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Csin%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%20%5Cleft(n%5Cin%5Cmathbb%7BN%7D%5Cright)%5C%5C%0Ab_n%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Ccos%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%5Cleft(n%5Cin%5Cmathbb%7BN%7D%5Cright)%0A%5Cend%7Baligned%7D$

則對任意 $n%5Cin%5Cmathbb%7BN%7D%5E*$ ，

$%5Cbegin%7Baligned%7D%0Aa_n%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Csin%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%20%5Cmathrm%7Bd%7D%20%5Cleft(-%5Ccos%20x%5Cright)%20%7D%5C%5C%26%3D-x%5En%5Ccos%20x%5Cbigg%7C_0%5E%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7B%5Cleft(-%5Ccos%20x%20%5Cright)%5Cmathrm%7Bd%7D%20x%5En%20%7D%5C%5C%26%3Dn%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5E%7Bn-1%7D%5Ccos%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%5C%5C%26%3Dnb_%7Bn-1%7D%5C%5C%0Ab_n%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%5Ccos%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5En%20%5Cmathrm%7Bd%7D%20%5Csin%20x%20%7D%5C%5C%26%3Dx%5En%5Csin%20x%5Cbigg%7C_0%5E%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7B%5Csin%20x%20%5Cmathrm%7Bd%7D%20x%5En%20%7D%5C%5C%26%3D%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5En-n%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%7Bx%5E%7Bn-1%7D%5Csin%20x%20%5Cmathrm%7Bd%7D%20x%20%7D%5C%5C%26%3D%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5En-na_%7Bn-1%7D%0A%5Cend%7Baligned%7D$

則對任意 $n%5Cgeq%202$ 且 $n%5Cin%5Cmathbb%7BN%7D$ ，

$%5Cbegin%7Baligned%7D%0Aa_n%26%3Dnb_%7Bn-1%7D%3Dn%5Ccdot%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-1%7D-n%5Cleft(n-1%5Cright)a_%7Bn-2%7D%5C%5C%0Ab_n%26%3D%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5En-na_%7Bn-1%7D%3D%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5En-n%5Cleft(n-1%5Cright)b_%7Bn-2%7D%0A%5Cend%7Baligned%7D$

當(dāng) $n%20%5Cto%20%5Cinfty$ 時，不斷迭代得

$%5Cbegin%7Baligned%7D%0Aa_n%26%3Dn%5Ccdot%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-1%7D-n%5Cleft(n-1%5Cright)%5Cleft(n-2%5Cright)%5Ccdot%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-3%7D%2B%5Ccdots%5C%5C%26%3D%5Csum_%7Bi%3D0%7D%5E%5Cinfty%7B%5Cfrac%7B%5Cleft(-1%5Cright)%5E%7Bi%7D%5Ccdot%20n!%7D%7B%5Cleft(n-2i-1%5Cright)!%7D%5Ccdot%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-2i-1%7D%7D%5C%5C%0Ab_n%26%3D%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5En-n%5Cleft(n-1%5Cright)%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-2%7D%2B%5Ccdots%5C%5C%26%3D%5Csum_%7Bi%3D0%7D%5E%5Cinfty%20%7B%5Cfrac%7B%5Cleft(-1%5Cright)%5E%7Bi%7D%5Ccdot%20n!%7D%7B%5Cleft(n-2i%5Cright)!%7D%5Ccdot%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5E%7Bn-2i%7D%7D%0A%5Cend%7Baligned%7D$

得

$L%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%20%20n%5E%7BC%2B1%7D%5Ccdot%5Cfrac%7Bn%5E%7B-1%7Da_n%7D%7Bb_n%7D%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Ccdot%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%20n%5E%7BC%2B1%7D$

若 $C%3C-1$ ，則

$L%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Ccdot%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%20n%5E%7BC%2B1%7D%3D0$

與 $L%20%5Cneq%200$ 矛盾，不符合題意.

若 $C%3E-1$ ，則

$L%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Ccdot%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%20n%5E%7BC%2B1%7D%3D%2B%5Cinfty$

與 $L%5Cin%5Cmathbb%7BR%7D$ 矛盾，不符合題意.

若 $C%3D-1$ ，則

$L%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Ccdot%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%20n%5E%7BC%2B1%7D%3D%5Cfrac%7B2%7D%7B%5Cpi%7D$

符合題意.

綜上， $C%3D-1%2CL%3D%5Cfrac%7B2%7D%7B%5Cpi%7D$ .

所以 $L%3D%5Cfrac%7B2%7D%7B%5Cpi%7D$ .

19. 已知 $n%20%5Cin%20%5Cmathbb%7BN%7D%5E*$ ，若存在正整數(shù) $a_1%2Ca_2%2C%5Ccdots%2Ca_n%2Cb_1%2Cb_2%2C%5Ccdots%2Cb_n$ ，滿足

$%5Csum_%7Bi%3D1%7D%5En%20a_i%5E2%5Ccdot%5Csum_%7Bi%3D1%7D%5En%20b_i%5E2-%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20a_i%20b_i%5Cright)%5E2%3Dn$

則符合條件的 $n$ 的個數(shù)為___________.

答案? 2

解析??

若 $n%3D1$ ，則

$%5Csum_%7Bi%3D1%7D%5En%20a_i%5E2%5Ccdot%5Csum_%7Bi%3D1%7D%5En%20b_i%5E2-%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20a_i%20b_i%5Cright)%5E2%3D0%20%5Cneq%201$

所以 $n%3D1$ 不符合題意.

若 $n%5Cgeq2$ ，由 $%5Ctext%7BLagrange%7D$ 恒等式

$n%3D%5Csum_%7Bi%3D1%7D%5En%20a_i%5E2%5Ccdot%5Csum_%7Bi%3D1%7D%5En%20b_i%5E2-%5Cleft(%5Csum_%7Bi%3D1%7D%5En%20a_i%20b_i%5Cright)%5E2%3D%5Csum_%7B1%5Cleq%20i%3Cj%5Cleq%20n%7D%7B%5Cleft(a_ib_j-a_jb_i%5Cright)%5E2%7D%20$

若 $n%3D2$ ，則

$%5Cleft(a_1b_2-a_2b_1%5Cright)%5E2%3D2$

即

$%5Cleft%7Ca_1b_2-a_2b_1%5Cright%7C%3D%5Csqrt%7B2%7D$

這是不可能成立的，所以 $n%3D2$ 不符合題意.

若 $n%5Cgeq%205$ ，設(shè) $%5Cfrac%7Bb_i%7D%7Ba_i%7D%5Cleft(1%5Cleq%20i%20%5Cleq%20n%5Cright)$ 的不同的值共有 $k%5Cleft(2%5Cleq%20k%20%5Cleq%20n%5Cright)$ 個，每個值對應(yīng)的 $i$ 的個數(shù)分別為 $x_1%2Cx_2%2C%5Ccdots%2Cx_k$ ，且

$x_j%20%5Cleq%20x_%7Bj%2B1%7D%5Cleft(1%5Cleq%20j%20%5Cleq%20k-1%5Cright)$

則

$%5Csum_%7Bj%3D1%7D%5Ek%20x_j%3Dn$

有 $1%5Cleq%20x_j%20%5Cleq%20n-k%2B1%E4%B8%94x_j%5Cin%20%5Cmathbb%7BN%7D%5E*%5Cleft(1%5Cleq%20j%5Cleq%20k%5Cright)$ .

由

$%5Csum_%7Bi%3D1%7D%5Ek%20%5Cleft(x_i-1%5Cright)%5Cleft(x_i-n%2Bk-1%5Cright)%5Cleq0$

得

$%5Csum_%7Bi%3D1%7D%5Ek%20x_i%5E2%3D%5Csum_%7Bi%3D1%7D%5Ek%20%5Cleft(x_i-1%5Cright)%5Cleft(x_i-n%2Bk-1%5Cright)%2Bn%5Cleft(n-k%2B2%5Cright)-k%5Cleft(n-k%2B1%5Cright)%5Cleq%20k%5E2-%5Cleft(2n%2B1%5Cright)k%2Bn%5E2%2B2n$

于是

$%5Csum_%7B1%5Cleq%20i%3Cj%5Cleq%20n%7D%7B%5Cleft(a_ib_j-a_jb_i%5Cright)%5E2%7D%20%3D%5Csum_%7B1%5Cleq%20i%3Cj%5Cleq%20n%7D%7Ba_i%5E2a_j%5E2%5Cleft(%5Cfrac%7Bb_j%7D%7Ba_j%7D-%5Cfrac%7Bb_i%7D%7Ba_i%7D%5Cright)%5E2%7D%5Cgeq%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5Ek%20x_i%5Cleft(n-x_i%5Cright)%7D%7B2%7D%3D%5Cfrac%7Bn%5E2-%5Csum_%7Bi%3D1%7D%5Ek%20x_i%5E2%20%7D%7B2%7D%5Cgeq%5Cfrac%7B-k%5E2%2B%5Cleft(2n%2B1%5Cright)k-2n%7D%7B2%7D$

由

$x_1%5Cleq%5Csum_%7Bj%3D2%7D%5Ek%20x_j%3Dn-x_1$

得 $x_1%5Cleq%5Cfrac%7Bn%7D%7B2%7D%3C%20n-2$ .

若 $x_1%20%5Cgeq%202$ ，則

矛盾，所以 $x_1%20%3C%202$ ，得 $x_1%20%3D1$ .

若 $k%3D2$ ，則 $%5Csum_%7B1%5Cleq%20i%3Cj%5Cleq%20n%7D%7B%5Cleft(a_ib_j-a_jb_i%5Cright)%5E2%7D$ 的組成為

$%5Cleft(%5Cfrac%7B1%7D%7B2%7Dn%5E2-%5Cfrac%7B3%7D%7B2%7Dn%2B1%5Cright)$ 個0， $%5Cleft(n-2%5Cright)$ 個1，1個2

但2不是完全平方數(shù)，因此該情況是不可能的，不符合題意.

若 $k%20%5Cgeq%203$ ，則

$%5Csum_%7B1%5Cleq%20i%3Cj%5Cleq%20n%7D%7B%5Cleft(a_ib_j-a_jb_i%5Cright)%5E2%7D%20%5Cgeq%5Cfrac%7B-k%5E2%2B%5Cleft(2n%2B1%5Cright)k-2n%7D%7B2%7D%20%5Cgeq%202n-3%20%3En$

矛盾，不符合題意.

所以 $n%5Cgeq%205$ 不符合題意.

若 $n%3D3$ ，取

$%5Cbegin%7Baligned%7D%0A%5Cleft(a_1%2Ca_2%2Ca_3%5Cright)%26%3D%5Cleft(1%2C2%2C1%5Cright)%5C%5C%0A%5Cleft(b_1%2Cb_2%2Cb_3%5Cright)%26%3D%5Cleft(3%2C5%2C2%5Cright)%0A%5Cend%7Baligned%7D$

符合題意.

若 $n%3D4$ ，取

$%5Cbegin%7Baligned%7D%0A%5Cleft(a_1%2Ca_2%2Ca_3%2Ca_4%5Cright)%26%3D%5Cleft(1%2C2%2C1%2C2%5Cright)%5C%5C%0A%5Cleft(b_1%2Cb_2%2Cb_3%2Cb_4%5Cright)%26%3D%5Cleft(2%2C3%2C2%2C3%5Cright)%0A%5Cend%7Baligned%7D$

符合題意.

綜上，符合條件的所有的 $n$ 為3, 4.

所以符合條件的 $n$ 的個數(shù)為2.

20. 已知 $%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Cln(1%2B%5Csin%5E2x)-6(%5Csqrt%5B3%5D%7B2-%5Ccos%20x%7D-1)%7D%7Bx%5E4%7D%7D%5Cright%7C%3D%5Cfrac%7Bq%7D%7Bp%7D$ ， $p$ 和 $q$ 是互素的正整數(shù)，則 $p%2Bq%3D$ ___________.

答案? 19

解析

$%5Cbegin%7Baligned%7D%0A%26%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Cln%5Cleft(1%2B%5Csin%5E2x%5Cright)-6%5Cleft(%5Csqrt%5B3%5D%7B2-%5Ccos%20x%7D-1%5Cright)%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Csin%5E2%20x-%5Cfrac%7B1%7D%7B2%7D%5Csin%5E4%20x%2B%5Comicron%5Cleft(%5Csin%5E4%20x%5Cright)-6%5Cleft(%5Csqrt%5B3%5D%7B1%2B2%5Csin%5E2%20%5Cfrac%7Bx%7D%7B2%7D%7D-1%5Cright)%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Cleft%5Bx-%5Cfrac%7B1%7D%7B6%7Dx%5E3%2B%5Comicron%5Cleft(x%5E4%5Cright)%5Cright%5D%5E2-%5Cfrac%7B1%7D%7B2%7Dx%5E4%20%2B%5Comicron%5Cleft(x%5E4%5Cright)-6%5Cleft%5B1%2B%5Cfrac%7B2%7D%7B3%7D%5Csin%5E2%20%5Cfrac%7Bx%7D%7B2%7D-%5Cfrac%7B4%7D%7B9%7D%5Csin%5E4%5Cfrac%7Bx%7D%7B2%7D%2B%5Comicron%5Cleft(%5Csin%5E4%5Cfrac%7Bx%7D%7B2%7D%5Cright)-1%5Cright%5D%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Cleft%5Bx-%5Cfrac%7B1%7D%7B6%7Dx%5E3%2B%5Comicron%5Cleft(x%5E4%5Cright)%5Cright%5D%5E2-%5Cfrac%7B1%7D%7B2%7Dx%5E4%20%2B%5Comicron%5Cleft(x%5E4%5Cright)-6%5Cleft%5B%5Cfrac%7B2%7D%7B3%7D%5Csin%5E2%20%5Cfrac%7Bx%7D%7B2%7D-%5Cfrac%7B4%7D%7B9%7D%5Csin%5E4%5Cfrac%7Bx%7D%7B2%7D%2B%5Comicron%5Cleft(%5Csin%5E4%5Cfrac%7Bx%7D%7B2%7D%5Cright)%5Cright%5D%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7Bx%5E2-%5Cfrac%7B1%7D%7B3%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)-%5Cfrac%7B1%7D%7B2%7Dx%5E4%20%2B%5Comicron%5Cleft(x%5E4%5Cright)-6%5Cleft%5B%5Cfrac%7B2%7D%7B3%7D%5Cleft%5B%5Cfrac%7B1%7D%7B2%7Dx-%5Cfrac%7B1%7D%7B48%7Dx%5E3%2B%5Comicron%5Cleft(%5Cfrac%7B1%7D%7B16%7Dx%5E4%5Cright)%5Cright%5D%5E2-%5Cfrac%7B1%7D%7B36%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)%5Cright%5D%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7Bx%5E2-%5Cfrac%7B5%7D%7B6%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)-6%5Cleft%5B%5Cfrac%7B1%7D%7B6%7Dx%5E2-%5Cfrac%7B1%7D%7B72%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)-%5Cfrac%7B1%7D%7B36%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)%5Cright%5D%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7Bx%5E2-%5Cfrac%7B5%7D%7B6%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)-6%5Cleft%5B%5Cfrac%7B1%7D%7B6%7Dx%5E2-%5Cfrac%7B1%7D%7B24%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)%5Cright%5D%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7Bx%5E2-%5Cfrac%7B5%7D%7B6%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)-x%5E2%2B%5Cfrac%7B1%7D%7B4%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B-%5Cfrac%7B7%7D%7B12%7Dx%5E4%2B%5Comicron%5Cleft(x%5E4%5Cright)%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C-%5Cfrac%7B7%7D%7B12%7D%2B%5Clim_%7Bx%20%5Crightarrow%200%7D%7B%5Cfrac%7B%5Comicron%5Cleft(x%5E4%5Cright)%7D%7Bx%5E4%7D%7D%5Cright%7C%5C%5C%0A%26%3D%5Cleft%7C-%5Cfrac%7B7%7D%7B12%7D%5Cright%7C%5C%5C%0A%26%3D%5Cfrac%7B7%7D%7B12%7D%0A%5Cend%7Baligned%7D$

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=p" alt="p">和 $q$ 是互素的正整數(shù)，所以 $p%3D12%2Cq%3D7$ .

因此 $p%2Bq%3D19$ .

21. 已知多項(xiàng)式 $f_n%5Cleft(x%5Cright)%2Cn%5Cgeq0$ ，滿足 $f_0%5Cleft(x%5Cright)%3D1$ ，當(dāng) $n%20%5Cgeq%201$ 時滿足 $f_n%5Cleft(0%5Cright)%3D0$ ；并且當(dāng) $n%20%5Cgeq%200$ 時有

$%5Cfrac%7B%5Cmathrm%7Bd%7Df_%7Bn%2B1%7D%5Cleft(x%5Cright)%7D%7B%5Cmathrm%7Bd%7Dx%7D%3D%5Cleft(n%2B1%5Cright)f_n%5Cleft(x%5Cright)$

請給出 $f_%7B100%7D%5Cleft(2023%5Cright)$ 在10進(jìn)制下的最后2位數(shù).

答案? 01.

解析

首先用數(shù)學(xué)歸納法證明 $f_n%5Cleft(x%5Cright)%3Dx%5En%5Cleft(x%5Cin%5Cmathbb%7BN%7D%5E*%5Cright)$ .

$%5Cfrac%7B%5Cmathrm%7Bd%7Df_%7B1%7D%5Cleft(x%5Cright)%7D%7B%5Cmathrm%7Bd%7Dx%7D%3D1%5Ccdot%20f_0%5Cleft(x%5Cright)%3D1$

得 $f_1%5Cleft(x%5Cright)%3Dx%2BC$ .

$f_1%5Cleft(0%5Cright)%3D0%3DC$ ，所以 $f_1%5Cleft(x%5Cright)%3Dx$ .

所以原命題對 $n%3D1$ 成立.

假設(shè)原命題對 $n%3Dk%5Cleft(k%5Cin%5Cmathbb%7BN%7D%5E*%5Cright)$ 成立，即

$f_k%5Cleft(x%5Cright)%3Dx%5Ek$

則

$%5Cfrac%7B%5Cmathrm%7Bd%7Df_%7Bk%2B1%7D%5Cleft(x%5Cright)%7D%7B%5Cmathrm%7Bd%7Dx%7D%3D%5Cleft(k%2B1%5Cright)f_k%5Cleft(x%5Cright)%3D%5Cleft(k%2B1%5Cright)x%5Ek$

得 $f_%7Bk%2B1%7D%5Cleft(x%5Cright)%3Dx%5E%7Bk%2B1%7D%2BC$ .

$f_%7Bk%2B1%7D%5Cleft(0%5Cright)%3D0%3DC$ ，所以 $f_1%5Cleft(x%5Cright)%3Dx%5E%7Bk%2B1%7D$ .

所以原命題對 $n%3Dk%2B1$ 成立.

所以原命題得證，即 $f_n%5Cleft(x%5Cright)%3Dx%5En%5Cleft(x%5Cin%5Cmathbb%7BN%7D%5E*%5Cright)$ .

$f_%7B100%7D%5Cleft(2023%5Cright)%3D2023%5E%7B100%7D%3D%5Cleft(2020%2B3%5Cright)%5E%7B100%7D%5Cequiv%203%5E%7B100%7D%3D9%5E%7B50%7D%3D%5Cleft(10-1%5Cright)%5E%7B50%7D%5Cequiv%20-500%2B1%5Cequiv%201%20%5Cpmod%20%7B100%7D$

所以 $f_%7B100%7D%5Cleft(2023%5Cright)$ 在10進(jìn)制下的最后2位數(shù)為01.

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