已知等比数列 $\left\{ {a_n} \right\}$ 的公比为 $q$，记 ${b_n} = {a_{m\left(n - 1\right) + 1}} + {a_{m\left(n - 1\right) + 2}} + \cdots + {a_{m\left(n - 1\right) + m}} $，$ {c_n} = {a_{m\left(n - 1\right) + 1}} \cdot {a_{m\left(n - 1\right) + 2}} \cdot \cdots \cdot {a_{m\left(n - 1\right) + m}} , \left( {m,n \in {{\mathbb{N}}^*}} \right)$，则以下结论一定正确的是 \((\qquad)\)
【难度】
【出处】
2013年高考福建卷（理）
【标注】
【答案】
C
【解析】
本题考查等比数列的判定，直接根据定义计算后一项比上前一项的值即可．$\because \left\{a_n\right\}$ 为等比数列．
$\begin{split}\therefore \dfrac{b_{n+1}}{b_n}&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{a_{m\left(n-1\right)+1}+a_{m\left(n-1\right)+2}+\cdots+a_{m\left(n-1\right)+m}}
\\&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{a_{mn+1-m}+a_{mn+2-m}+\cdots+a_{mn+m-m}}
\\&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{q^{-m}\left(a_{mn+1}+a_{mn+2}+\cdots+a_{mn+m}\right)}\\&=q^m ,\end{split}$
$\dfrac{c_{n+1}}{c_n}=\dfrac{a_{mn+1}\cdot a_{mn+2}\cdot \cdots \cdot a_{mn+m}}{a_{m\left(n-1\right)+1}\cdot a_{m\left(n-1\right)+2}\cdot \cdots\cdot a_{m\left(n-1\right)+m}}
=\left(q^m\right)^m=q^{m^2}$．
即 $\left\{b_n\right\}$ 为等比数列，公比为 $q^m$；$\left\{c_n\right\}$ 为等比数列，公比为 $q^{m^2}$．
$\begin{split}\therefore \dfrac{b_{n+1}}{b_n}&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{a_{m\left(n-1\right)+1}+a_{m\left(n-1\right)+2}+\cdots+a_{m\left(n-1\right)+m}}
\\&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{a_{mn+1-m}+a_{mn+2-m}+\cdots+a_{mn+m-m}}
\\&=\dfrac{a_{mn+1}+a_{mn+2}+\cdots +a_{mn+m}}{q^{-m}\left(a_{mn+1}+a_{mn+2}+\cdots+a_{mn+m}\right)}\\&=q^m ,\end{split}$
$\dfrac{c_{n+1}}{c_n}=\dfrac{a_{mn+1}\cdot a_{mn+2}\cdot \cdots \cdot a_{mn+m}}{a_{m\left(n-1\right)+1}\cdot a_{m\left(n-1\right)+2}\cdot \cdots\cdot a_{m\left(n-1\right)+m}}
=\left(q^m\right)^m=q^{m^2}$．
即 $\left\{b_n\right\}$ 为等比数列，公比为 $q^m$；$\left\{c_n\right\}$ 为等比数列，公比为 $q^{m^2}$．
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