909

在正项等比数列 ${\displaystyle \{a_n\}}$ 中，${\displaystyle a_5={\displaystyle \frac{1}{2},a_6+a_7=3}}$, 则满足 ${\displaystyle a_1+a_2+\cdots +a_n>a_1{a}_2\cdots a_n}$ 的最大正整数 ${\displaystyle n}$ 的值为_____.

首先注意到$$\begin{array}{r}{A}_6+a_7=a_5(q+q^2)=\frac{1}{2}(q+q^2)=3,\end{array}$$
并且在正项等比数列中 ${\displaystyle q>0}$，因此解得 ${\displaystyle q=2}$，进而 ${\displaystyle a_n=2^{n-6}}$. 从而$$\begin{array}{r}0<a_1+\cdots +a_n-a_1\cdots a_n=\frac{1}{32}(2^n-1)-2^{\frac{(-5+n-6)n}{2}}=\frac{1}{32}(2^n-2^{\frac{n^2-11n+10}{2}}-1).\end{array}$$
只需要$$\begin{array}{r}{2}^n-1\ge 2^{n-1}>2^{\frac{n^2-11n+10}{2}}\Rightarrow n-1\ge \frac{n^2-11n+10}{2},\end{array}$$
解得 ${\displaystyle 1\le n\le 12}$, 即 ${\displaystyle n_{max}=12}$.