2023浙江大学强基数学逐题解析（6）

2023-06-24 21:59 作者:CHN_ZCY 0人读过 | 我要投稿

封面：《不当哥哥了》

14. 已知数列 $%5Cleft%5C%7Ba_n%5Cright%5C%7D$ 中有 $a_1%3D%5Cfrac%7B1%7D%7B2%7D$ ，

$a_%7Bn%2B1%7D%3D%5Cfrac%7Ba_n%7D%7B%5Cleft(1-%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7Da_n%2B%5Csqrt%7B2%7D%2B1%7D%5Cleft(n%5Cin%5Cmathbb%7BN%7D%5E*%5Cright)$

则 $%5Clim_%7Bn%5Cto%2B%5Cinfty%7D%20%5Csqrt%5Bn%5D%7Ba_n%7D%3D$ ___________.

答案 $%5Csqrt%7B2%7D-1$

解析

$%5Cfrac%7B1%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cleft(1-%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D%2B%5Cfrac%7B%5Csqrt%7B2%7D%2B1%7D%7Ba_n%7D$

$%5Cfrac%7B1%7D%7B%5Cleft(%5Csqrt%7B2%7D%2B1%5Cright)%5E%7Bn%2B1%7Da_%7Bn%2B1%7D%7D%3D%5Cleft(2%5Csqrt%7B2%7D-3%5Cright)%5E%7Bn%2B1%7D%2B%5Cfrac%7B1%7D%7B%5Cleft(%5Csqrt%7B2%7D%2B1%5Cright)%5Ena_n%7D$

所以

$%5Cbegin%7Baligned%7D%0A%5Cfrac%7B1%7D%7B%5Cleft(%5Csqrt%7B2%7D%2B1%5Cright)%5E%7Bn%7Da_%7Bn%7D%7D%26%3D%5Cfrac%7B1%7D%7B%5Cleft(%5Csqrt%7B2%7D%2B1%5Cright)%5E1a_1%7D%2B%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%5Cleft(2%5Csqrt%7B2%7D-3%5Cright)%5Ei%20%5C%5C%26%0A%3D2%5Csqrt%7B2%7D-2%2B%5Cfrac%7B%5Cleft(%5Csqrt%7B2%7D-1%5Cright)%5E3%5Cleft%5B1-%5Cleft(2%5Csqrt%7B2%7D-3%5Cright)%5E%7Bn-1%7D%5Cright%5D%7D%7B2%5Csqrt%7B2%7D%7D%0A%5Cend%7Baligned%7D$

得

$a_n%3D%5Cfrac%7B2%5Csqrt%7B2%7D%7D%7B%5Cleft(1%2B%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D-%5Cleft(1-%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D%7D$

所以

$%5Cbegin%7Baligned%7D%0A%5Clim_%7Bn%5Cto%2B%5Cinfty%7D%7B%5Csqrt%5Bn%5D%7Ba_n%7D%7D%26%3D%5Clim_%7Bn%5Cto%2B%5Cinfty%7D%7B%5Cleft%5B%5Cfrac%7B2%5Csqrt%7B2%7D%7D%7B%5Cleft(1%2B%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D-%5Cleft(1-%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%7D%5C%5C%26%0A%3D%5Clim_%7Bn%5Cto%2B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft%5B%5Cleft(1%2B%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D-%5Cleft(1-%5Csqrt%7B2%7D%5Cright)%5E%7Bn%2B1%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%7D%7D%5C%5C%26%0A%3D%5Clim_%7Bn%5Cto%2B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft(1%2B%5Csqrt%7B2%7D%5Cright)%5Cleft%5B1-%5Cleft(2%5Csqrt%7B2%7D-3%5Cright)%5E%7Bn%2B1%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%7D%7D%5C%5C%26%0A%3D%5Cfrac%7B1%7D%7B1%2B%5Csqrt%7B2%7D%7D%5C%5C%26%0A%3D%5Csqrt%7B2%7D-1%0A%0A%5Cend%7Baligned%7D$

15.

的值为___________.

答案 $%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D$

解析

$%5Cbegin%7Baligned%7D%0A%5Ctan12%5E%5Ccirc%5Cleft(1%2B%5Cfrac%7B1%7D%7B%5Csin6%5E%5Ccirc%7D%5Cright)%26%3D%5Cfrac%7B1%7D%7B%5Csin6%5E%5Ccirc%7D%5Ccdot%5Ctan12%5E%5Ccirc%5Cleft(%5Csin6%5E%5Ccirc%2B1%5Cright)%5C%5C%26%3D%5Cfrac%7B1%7D%7B%5Csin6%5E%5Ccirc%7D%5Ccdot%5Ctan12%5E%5Ccirc%5Cleft(1-%5Ccos96%5E%5Ccirc%5Cright)%5C%5C%26%3D%5Cfrac%7B%5Csin96%5E%5Ccirc%7D%7B%5Csin6%5E%5Ccirc%7D%5Ccdot%5Ctan12%5E%5Ccirc%5Ccdot%5Cfrac%7B1-%5Ccos96%5E%5Ccirc%7D%7B%5Csin96%5E%5Ccirc%7D%5C%5C%26%3D%5Cfrac%7B%5Ccos6%5E%5Ccirc%7D%7B%5Csin6%5E%5Ccirc%7D%5Ccdot%5Ctan12%5E%5Ccirc%5Ccdot%5Ctan48%5E%5Ccirc%5C%5C%26%3D%5Cfrac%7B%5Ctan12%5E%5Ccirc%5Ccdot%5Ctan48%5E%5Ccirc%7D%7B%5Ctan6%5E%5Ccirc%7D%5C%5C%26%3D%5Cfrac%7B%5Ctan12%5E%5Ccirc%5Ctan48%5E%5Ccirc%5Ctan54%5E%5Ccirc%5Ctan66%5E%5Ccirc%5Ctan72%5E%5Ccirc%7D%7B%5Ctan6%5E%5Ccirc%5Ctan54%5E%5Ccirc%5Ctan66%5E%5Ccirc%5Ctan72%5E%5Ccirc%7D%5C%5C%26%3D%5Cfrac%7B%5Ctan36%5E%5Ccirc%5Ctan54%5E%5Ccirc%5Ctan66%5E%5Ccirc%7D%7B%5Ctan18%5E%5Ccirc%5Ctan72%5E%5Ccirc%7D%5C%5C%26%3D%5Ctan66%5E%5Ccirc%0A%5Cend%7Baligned%7D$

所以

$%5Cfrac%7B%5Ctan96%5E%5Ccirc-%5Ctan12%5E%5Ccirc%5Cleft(1%2B%5Cfrac%7B1%7D%7B%5Csin6%5E%5Ccirc%7D%5Cright)%7D%7B1%2B%5Ctan96%5E%5Ccirc%5Ctan12%5E%5Ccirc%5Cleft(1%2B%5Cfrac%7B1%7D%7B%5Csin6%5E%5Ccirc%7D%5Cright)%7D%3D%5Cfrac%7B%5Ctan96%5E%5Ccirc-%5Ctan66%5E%5Ccirc%7D%7B1%2B%5Ctan96%5E%5Ccirc%5Ctan66%5E%5Ccirc%7D%3D%5Ctan30%5E%5Ccirc%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D$

16. 已知数列 $%5Cleft%5C%7Ba_n%5Cright%5C%7D$ 满足 $a_1%3D1$ ， $a_%7Bn%2B1%7D%3D%5Cfrac%7Ba_n%7D%7B2a_n%5E2%2B1%7D$ ，则 $%5Cleft%5B2%5Clg%20a_%7B2023%7D%5Cright%5D%3D$ ___________.

答案 $-4$

解析

由 $a_%7Bn%2B1%7D%3D%5Cfrac%7Ba_n%7D%7B2a_n%5E2%2B1%7D$ ，即

$%5Cfrac%7B1%7D%7Ba_%7Bn%2B1%7D%7D%3D2a_n%2B%5Cfrac%7B1%7D%7Ba_n%7D$

即

$%5Cfrac%7B1%7D%7Ba_%7Bn%2B1%7D%5E2%7D%3D4a_n%5E2%2B%5Cfrac%7B1%7D%7Ba_n%5E2%7D%2B4$

所以

$%5Cfrac%7B1%7D%7Ba_%7Bn%7D%5E2%7D%3D4%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7Ba_i%5E2%7D%2B%5Cfrac%7B1%7D%7Ba_1%5E2%7D%2B4%5Cleft(n-1%5Cright)%3D4n-3%2B4%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7Ba_i%5E2%7D$

因此

$%5Cfrac%7B1%7D%7Ba_%7Bn%7D%5E2%7D%3D4n-3%2B4%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7Ba_i%5E2%7D%5Cgeq4n-3$

即

$a_n%5E2%5Cleq%5Cfrac%7B1%7D%7B4n-3%7D$

当且仅当 $n%3D1$ 时取等.

当 $n%5Cgeq2%E4%B8%94n%5Cin%5Cmathbb%7BN%7D%5E*$ 时，

$%5Cbegin%7Baligned%7D%0A%5Cfrac%7B1%7D%7Ba_%7Bn%7D%5E2%7D%26%3D4n-3%2B4%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7Ba_i%5E2%7D%5C%5C%26%3C4n-3%2B4%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7B%5Cfrac%7B1%7D%7B4i-3%7D%7D%5C%5C%26%3D4n-3%2B%5Csum_%7Bi%3D1%7D%5E%7Bn-1%7D%7B%5Cfrac%7B1%7D%7Bi-%5Cfrac%7B3%7D%7B4%7D%7D%7D%5C%5C%26%3C4n-3%2B4%2B%5Csum_%7Bi%3D2%7D%5E%7Bn-1%7D%7B%5Cln%5Cfrac%7Bi-%5Cfrac%7B3%7D%7B4%7D%7D%7Bi-%5Cfrac%7B7%7D%7B4%7D%7D%7D%5C%5C%26%3D4n%2B1%2B%5Cln%5Cleft(4n-7%5Cright)%0A%5Cend%7Baligned%7D$

即

$a_n%5E2%3E%5Cfrac%7B1%7D%7B4n%2B1%2B%5Cln%5Cleft(4n-7%5Cright)%7D$

所以当 $n%5Cgeq2%E4%B8%94n%5Cin%5Cmathbb%7BN%7D%5E*$ 时，

$%5Cfrac%7B1%7D%7B4n%2B1%2B%5Cln%5Cleft(4n-7%5Cright)%7D%3Ca_n%5E2%3C%5Cfrac%7B1%7D%7B4n-3%7D$

所以

$10%5E%7B-4%7D%3C%5Cfrac%7B1%7D%7B8093%2B%5Cln%5Cleft(8085%5Cright)%7D%3Ca_%7B2023%7D%5E2%3C%5Cfrac%7B1%7D%7B8089%7D%3C10%5E%7B-3%7D$

因此

$-4%3C2%5Clg%20a_%7B2023%7D%3C-3$

得 $%5Cleft%5B2%5Clg%20a_%7B2023%7D%5Cright%5D%3D-4$ .

17. 已知正的十进制五位数 $n$ 满足 $2556%5Cmid%20n%5E3-1$ ，则 $n$ 在十进制下的各位数字之和的最小值为___________.

答案 7

解析

因为 $2556%3D2%5E2%5Ccdot3%5E2%5Ccdot71%5E1$ ，所以

$%5Cbegin%7Baligned%7D%0A2556%5Cmid%20n%5E3-1%20%26%5CLeftrightarrow%20%0A%5Cbegin%7Bcases%7D%0A4%20%5Cmid%20n%5E3-1%5C%5C%0A9%20%5Cmid%20n%5E3-1%5C%5C%0A71%20%5Cmid%20n%5E3-1%0A%5Cend%7Bcases%7D%0A%5C%5C%26%5CLeftrightarrow%20%0A%5Cbegin%7Bcases%7D%0An%5E3%20%5Cequiv%201%20%5Cpmod%204%20%5C%5C%0An%5E3%20%5Cequiv%201%20%5Cpmod%209%5C%5C%0An%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D%0A%5Cend%7Bcases%7D%0A%5C%5C%26%5CLeftrightarrow%20%0A%5Cbegin%7Bcases%7D%0An%20%5Cequiv%201%20%5Cpmod%204%20%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%203%5C%5C%0An%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D%0A%5Cend%7Bcases%7D%0A%5Cend%7Baligned%7D$

由于 $n%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ ，所以 $%5Cleft(n%5E3%2C71%5Cright)%3D1$ .

由于71是质数，根据Fermat小定理， $n%5E%7B70%7D%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ .

因为 $n%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ ，所以 $%5Cdelta_%7B71%7D%5Cleft(n%5Cright)%5Cmid3$ .

因为 $n%5E%7B70%7D%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ ，所以 $%5Cdelta_%7B71%7D%5Cleft(n%5Cright)%5Cmid70$ .

因此 $%5Cdelta_%7B71%7D%5Cleft(n%5Cright)%5Cmid%5Cleft(3%2C70%5Cright)$ .

所以 $n%3Dn%5E%7B%5Cleft(3%2C70%5Cright)%7D%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ .

另一方面，当 $n%5Cequiv%201%20%5Cpmod%20%7B71%7D$ 时， $n%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D$ .

所以

$%5Cbegin%7Bcases%7D%0An%20%5Cequiv%201%20%5Cpmod%204%20%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%203%5C%5C%0An%5E3%20%5Cequiv%201%20%5Cpmod%20%7B71%7D%0A%5Cend%7Bcases%7D%0A%5CLeftrightarrow%0A%5Cbegin%7Bcases%7D%0An%20%5Cequiv%201%20%5Cpmod%204%20%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%203%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%20%7B71%7D%0A%5Cend%7Bcases%7D$

由中国剩余定理得

$%5Cbegin%7Bcases%7D%0An%20%5Cequiv%201%20%5Cpmod%204%20%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%203%5C%5C%0An%20%5Cequiv%201%20%5Cpmod%20%7B71%7D%0A%5Cend%7Bcases%7D%0A%5CLeftrightarrow%0An%5Cequiv1%5Cpmod%7B852%7D$