设正整数 n=a0⋅20+a1⋅2+⋯+ak−1⋅2k−1+ak⋅2kn = {a}_{0} \cdot {2}^{0} + {a}_{1} \cdot 2 + \cdots + {a}_{k - 1} \cdot {2}^{k - 1} + {a}_{k} \cdot {2}^{k}n=a0⋅20+a1⋅2+⋯+ak−1⋅2k−1+ak⋅2k ,其中 ai∈{0,1}{a}_{i} \in \{ 0,1\}ai∈{0,1} ,记 ω(n)=a0+a1+⋯+ak\omega \left( n\right) = {a}_{0} + {a}_{1} + \cdots + {a}_{k}ω(n)=a0+a1+⋯+ak . 则 ( )A. ω(2n)=ω(n)\omega \left( {2n}\right) = \omega \left( n\right)ω(2n)=ω(n)B. ω(2n+3)=ω(n)+1\omega \left( {{2n} + 3}\right) = \omega \left( n\right) + 1ω(2n+3)=ω(n)+1C. ω(8n+5)=ω(4n+3)\omega \left( {{8n} + 5}\right) = \omega \left( {{4n} + 3}\right)ω(8n+5)=ω(4n+3)D. ω(2n−1)=n\omega \left( {{2}^{n} - 1}\right) = nω(2n−1)=n