$A.$ $\left\{\mathrm{S}_{\mathrm{n}}\right\}$ 为递减数列 $B.$ $\left\{\mathrm{S}_{\mathrm{n}}\right\}$ 为递增数列 $C.$ $\left\{\mathbf{S}_{2 n-1}\right\}$ 为递增数列, $\left\{\mathbf{S}_{2 n}\right\}$ 为递减数列 $D.$ $\left\{\mathrm{S}_{2 \mathrm{n}-1}\right\}$ 为递减数列, $\left\{\mathrm{S}_{2 \mathrm{n}}\right\}$ 为递增数列

B

#### 解析：

【分析】由 $a_{n+1}=a_{n}$ 可知 $\triangle A_{n} B_{n} C_{n}$ 的边 $B_{n} C_{n}$ 为 定值 $a_{1}$, 由 $b_{n+1}+c_{n+1}-2 a_{1}=$
$\frac{1}{2}\left(b_{n}+c_{n}-2 a_{1}\right)$ 及 $b_{1}+c_{1}=2 a_{1}$ 得 $b_{n}+c_{n}=2 a_{1}$, 则在 $\triangle A_{n} B_{n} C_{n}$ 中边长 $B_{n} C_{n}=a_{1}$ 为定

$$\therefore b_{1}-a_{1}=2 a_{1}-c_{1}-a_{1}=a_{1}-c_{1} > 0, \quad \therefore b_{1} > a_{1} > c_{1} \text {, }$$

$\therefore b_{n}+c_{n}-2 a_{n}=0, \quad \therefore b_{n}+c_{n}=2 a_{n}=2 a_{1}, \quad \therefore b_{n}+c_{n}=2 a_{1}$,

$\therefore b_{n+1}-a_{1}=\frac{1}{2}\left(a_{1}-b_{n}\right), \quad \therefore b_{n}-a_{1}=\left(-\frac{1}{2}\right)^{n-1}$,
$\therefore b_{n}=a_{1}+\left(b_{1}-a_{1}\right)\left(-\frac{1}{2}\right)^{r-1}, \quad c_{n}=2 a_{1}-b_{n}=a_{1}-\left(b_{1}-a_{1}\right)\left(-\frac{1}{2}\right)^{n-1}$,
$\therefore \quad S_{n}^{2}=\frac{3 a_{1}}{2}\left(\frac{3 a_{1}}{2}-a_{1}\right)\left[\frac{3 a_{1}}{2}-a_{1}-\left(b_{1}-a_{1}\right)\left(-\frac{1}{2}\right)^{\mathrm{n}-1}\right][$
$\frac{3 a_{1}}{2}-a_{1}+\left(b_{1}-a_{1}\right)\left(-\frac{1}{2}\right)^{n-1]}$
$=\frac{3}{4} a_{1}^{2}\left[\frac{a_{1}^{2}}{2}-\left(\frac{1}{4}\right)^{n-1}\left(b_{1}-a_{1}\right)^{2}\right]$ 单调递增（可证当 $n=1$ 时 $\frac{a_{1}^{2}}{4}-\left(b_{1}-a_{1}\right)^{2} > 0$
)