927

已知等比数列 ${\displaystyle \{a_n\}}$, 公比为 ${\displaystyle q}$, 前 ${\displaystyle n}$ 项和为 ${\displaystyle S_n}$, ${\displaystyle S_2=4,S_5=16}$, 则下列说法不正确的是

• A. ${\displaystyle a_1>0}$
• B. ${\displaystyle 0<|q|<2}$
• C. ${\displaystyle S_n>0}$
• D. ${\displaystyle \frac{S_n}{n}}$ 一定有最小值

由题意$$\begin{array}{r}4=\frac{S_5}{S_2}=\frac{1-q^5}{1-q^2}=\frac{1+q+q^2+q^3+q^4}{1+q}\Rightarrow q^4+q^3+q^2-3q-3=0.\end{array}$$
记 ${\displaystyle f(x)=x^4+x^3+x^2-3x-3}$, 则 ${\displaystyle f^{\prime }(x)=4x^3+3x^2+2x-3,f^{″}(x)=12x^2+6x+2>0}$，故 ${\displaystyle f^{\prime }(x)}$ 存在唯一零点，${\displaystyle f(x)}$ 存在唯一极小值点. 又 ${\displaystyle f(-1)>0>f(0),f(1)<0<f(2)}$，故 ${\displaystyle q}$ 有解 ${\displaystyle q_1\in (-1,0),q_2\in (1,2)}$.

• A: 由于 ${\displaystyle q+1>0}$, 故 ${\displaystyle a_1={\displaystyle \frac{S_2}{1+q}>0}}$, A 正确.
• B: ${\displaystyle |q_1|,|q_2|\in (0,2)}$，B 正确.
• C: ${\displaystyle {\displaystyle S_n=\frac{a_1(1-q^n)}{1-q}}}$. 对 ${\displaystyle q_1}$, ${\displaystyle 1-q_1^n,1-q_1>0}$; 对 ${\displaystyle q_2}$，${\displaystyle 1-q_2^n,1-q_2>0}$. 故 C 正确.
• D: 取 ${\displaystyle q=q_1<0}$. 这样 ${\displaystyle a_{2n}<0\Rightarrow S_{2n-1}>S_{2n}>0\Rightarrow {\displaystyle \frac{S_{2n-1}}{2n-1}>\frac{S_{2n}}{2n}}}$. 这样 ${\displaystyle {\displaystyle \left\{\frac{S_n}{n}\right\}}}$ 的最小值若存在，只会在子列 ${\displaystyle {\displaystyle \left\{\frac{S_{2n}}{2n}\right\}}}$ 上取到. 注意到$$\begin{array}{rl}\frac{S_{2n+2}}{2n+2}-\frac{S_{2n}}{2n}& =\frac{a_1}{1-q}(\frac{1-q^{2n+2}}{2n+2}-\frac{1-q^{2n}}{2n})\\ & =\frac{a_1(1+q)q^{2n}}{n(n+2)}(n-\frac{1}{q^{2n}}-\frac{1}{q^{2n-2}}-\cdots -\frac{1}{q^2})<0,\end{array}$$ 因此 ${\displaystyle q=q_1}$ 时 ${\displaystyle {\displaystyle \left\{\frac{S_{2n}}{2n}\right\}}}$ 严格单减，${\displaystyle {\displaystyle \left\{\frac{S_n}{n}\right\}}}$ 没有最小值，D错.

因此这道题选 D.