$a_n=3\cdot 2^{n-1}$．

将条件 $\lambda=0$，$\mu=-2$ 代入可分析知 $\{a_n\}$ 是一个等比数列．由 $\lambda=0$，$\mu=-2$，有 $a_{n+1}a_n=2a_n^2\left(n\in {\mathbb{N}}_+\right)$．
若存在某个 $n_0\in {\mathbb{N}}_+$，使得 $a_{n_0}=0$，则由上述递推公式易得 $a_{n_0-1}=0$，重复上述过程可得 $a_1=0$，此与 $a_1=3$ 矛盾，所以对任意 $n\in {\mathbb{N}}_+$，$a_n\neq 0$．
从而 $a_{n+1}=2a_n\left(n\in {\mathbb{N}}_+\right)$，即 $\left\{a_n\right\}$ 是一个公比 $q=2$ 的等比数列．
故 $a_n=a_1q^{n-1}=3\cdot 2^{n-1}$．

略．

可先由递推关系分析出 $\{a_n\}$ 的单调性和上界，然后计算 $a_{n+1}$ 和 $a_n$ 的差，继而用累加法求 $a_{{k_0}+1}$，然后通过 $a_n$ 的上界 $3$ 放缩，从而证得不等式左边成立．然后利用 $a_n$ 的下界 $2$ 放缩，证得右边不等式成立．由 $\lambda=\dfrac{1}{k_0}$，$\mu=-1$，数列 $\left\{a_n\right\}$ 的递推关系式变为 $a_{n+1}a_n+\dfrac{1}{k_0}a_{n+1}-a_n^2=0$，变形为 $a_{n+1}\left(a_n+\dfrac{1}{k_0}\right)=a_n^2\left(n\in {\mathbb{N}}_+\right)$．
由上式及 $a_1=3>0$，归纳可得 $3=a_1>a_2>\cdots>a_n>a_{n+1}>\cdots>0$．
因为 $a_{n+1}=\dfrac{a_n^2}{a_n+\dfrac{1}{k_0}}=\dfrac{a_n^2-\dfrac{1}{k_0^2}+\dfrac{1}{k_0^2}}{a_n+\dfrac{1}{k_0}}=a_n-\dfrac{1}{k_0}+\dfrac{1}{k_0}\cdot \dfrac{1}{k_0a_n+1}$，
所以对 $n=1$，$2$，$\cdots$，$k_0$ 求和得\[ \begin{split}a_{k_0+1}&=a_1+\left(a_2-a_1\right)+\cdots+\left(a_{k_0+1}-a_{k_0}\right)\\&=a_1-k_0\cdot\dfrac{1}{k_0}+\dfrac{1}{k_0}\cdot\left(\dfrac{1}{k_0a_1+1}+\dfrac{1}{k_0a_2+1}+\cdots+\dfrac{1}{k_0a_{k_0}+1}\right)\\& >2+\dfrac{1}{k_0}\cdot\left(\underbrace{\dfrac{1}{3k_0+1}+\dfrac{1}{3k_0+1}+\cdots+\dfrac{1}{3k_0+1}}_{k_0个}\right)\\&=2+\dfrac{1}{3k_0+1} .\end{split} \]另一方面，由上已证的不等式知 $a_1>a_2>\cdots>a_{k_0}>a_{k_0+1}>2$，得\[ \begin{split}a_{k_0+1}&=a_1-k_0\cdot\dfrac{1}{k_0}+\dfrac{1}{k_0}\cdot\left(\dfrac{1}{k_0a_1+1}+\dfrac{1}{k_0a_2+1}+\cdots+\dfrac{1}{k_0a_k+1}\right)\\& <2+\dfrac{1}{k_0}\cdot \left(\underbrace{\dfrac{1}{2k_0+1}+\dfrac{1}{2k_0+1}+\cdots+\dfrac{1}{2k_0+1}}_{k_0个}\right)\\&=2+\dfrac{1}{2k_0+1} .\end{split} \]综上，$2+\dfrac{1}{3k_0+1}<a_{k_0+1}<2+\dfrac{1}{2k_0+1}$．