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2021年高考数学新高考Ⅱ-11<-->2021年高考数学新高考Ⅱ-13
12．（5分）设正整数$n=a_{0}\cdot 2^{0}+a_{1}\cdot 2^{1}+\ldots +a_{k-1}\cdot 2^{k-1}+a_{k}\cdot 2^{k}$，其中$a_{i}\in \{0$，$1\}$，记$\omega (n)=a_{0}+a_{1}+\ldots +a_{k}$，则( )
A．$\omega (2n)=\omega (n)$
B．$\omega (2n+3)=\omega (n)+1$
C．$\omega (8n+5)=\omega (4n+3)$
D．$\omega (2^{n}-1)=n$
分析：$2n=a_{0}\cdot 2^{1}+a_{1}\cdot 2^{2}+\ldots +a_{k-1}\cdot 2^{k}+a_{k}\cdot 2^{k+1}$可判断$A$；取$n=2$可判断$B$；
把$8n+5$和$4n+3$都化成$n=a_{0}\cdot 2^{0}+a_{1}\cdot 2^{1}+\ldots +a_{k-1}\cdot 2^{k-1}+a_{k}\cdot 2^{k}$，可判断$C$；
$2^{n}-1=1\cdot 2^{0}+1\cdot 2^{1}+\cdot \cdot \cdot +1\cdot 2^{n-1}$可判断$D$．
解：$\because 2n=a_{0}\cdot 2^{1}+a_{1}\cdot 2^{2}+\ldots +a_{k-1}\cdot 2^{k}+a_{k}\cdot 2^{k+1}$，$\therefore \omega (2n)=\omega (n)=a_{0}+a_{1}+\ldots +k$，$\therefore A$对；
当$n=2$时，$2n+3=7=1\cdot 2^{0}+1\cdot 2^{1}+1\cdot 2^{2}$，$\therefore \omega$（7）$=3$．$\because 2=0\cdot 2^{0}+1\cdot 2^{1}$，$\therefore \omega$（2）$=0+1=1$，$\therefore \omega$（7）$\ne \omega$（2）$+1$，$\therefore B$错；
$\because 8n+5=a_{0}\cdot 2^{3}+a_{1}\cdot 2^{4}+\cdot \cdot \cdot +a_{k}\cdot 2^{k+3}+5=1\cdot 2^{0}+1\cdot 2^{2}+a_{0}\cdot 2^{3}+a_{1}\cdot 2^{4}+\cdot \cdot \cdot +a_{k}\cdot 2^{k+3}$，
$\therefore \omega (8n+5)=a_{0}\cdot +a_{1}\cdot +\cdot \cdot \cdot +a_{k}+2$．$\because 4n+3=a_{0}\cdot 2^{2}+a_{1}\cdot 2^{3}+\cdot \cdot \cdot +a_{k}\cdot 2^{k+2}+3=1\cdot 2^{0}+1\cdot 2^{1}+a_{0}\cdot 2^{2}+a_{1}\cdot 2^{3}+\cdot \cdot \cdot +a_{k}\cdot 2^{k+2}$，
$\therefore \omega (4n+3)=a_{0}\cdot +a_{1}\cdot +\cdot \cdot \cdot +a_{k}+2=\omega (8n+5)$．$\therefore C$对；
$\because 2^{n}-1=1\cdot 2^{0}+1\cdot 2^{1}+\cdot \cdot \cdot +1\cdot 2^{n-1}$，$\therefore \omega (2^{n}-1)=n$，$\therefore D$对．
故选：$ACD$．
点评：本题考查数列递推式，考查数学运算能力，属于难题．
2021年高考数学新高考Ⅱ-11<-->2021年高考数学新高考Ⅱ-13
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