点 $D$ 为 $\triangle ABC$ 边 $BC$ 上一点，$\overrightarrow{BC}=4\overrightarrow {DC}$，$E_n$（$n\in\mathbb N^{\ast}$）为边 $AC$ 上的点列，且满足 $\overrightarrow{E_nA}=\dfrac14a_{n+1}\overrightarrow{E_nB}-\left(3a_n+3^{n+1}\right)\overrightarrow{E_nD}$，若 $a_1=3$，则 $a_n=$_______．
解 $n\cdot 3^n$．
根据换底公式，有\[\begin{split}\left(\dfrac 14a_{n+1}-3a_n-3^{n+1}-1\right)\overrightarrow{AE_n}&=\dfrac 14a_{n+1}\overrightarrow{AB}-\left(3a_n+3^{n+1}\right)\overrightarrow{AD}\\ &=\dfrac 14a_{n+1}\overrightarrow{AB}-\left(3a_n+3^{n+1}\right)\left(\dfrac 14\overrightarrow{AB}+\dfrac 34\overrightarrow{AC}\right)\\ &=\left(\dfrac 14a_{n+1}-\dfrac 34a_n-\dfrac {3^{n+1}}4\right)\overrightarrow{AB}-\left(\dfrac 94a_n+\dfrac{3^{n+2}}4\right)\overrightarrow{AC},\end{split}\]于是\[a_{n+1}=3a_n+3^{n+1},\]也即\[\dfrac{a_{n+1}}{3^{n+1}}=\dfrac{a_n}{3^n}+1,\]因此\[a_n=n\cdot 3^n,n\in\mathbb N^{\ast}.\]