Sejam $$\mathrm{A}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\left[\begin{array}{l}\mathrm{n}\\ \mathrm{k}\end{array}\right]\cdot 3^{\mathrm{k}}\mathrm{e}\mathrm{B}=\sum _{k=0}^{n-1}\left[\begin{array}{c}n-1\\ k\end{array}\right]\cdot 11^k$$Se $\ln B-\ln A=ln\frac{6561}{4}$então $n$ é igual a: