If  $\mathrm{L}=\underset{\mathrm{x}\to \infty }{\lim } {\left(2\times 3^2\times 2^3\times 3^4...\times 2^{\mathrm{n}-1}\times 3^{\mathrm{n}}\right)}^{\frac{1}{\left({\mathrm{n}}^2+1\right)}}$, then the value of L⁴   is _____.

$$\begin{gathered} \mathrm{Clearly}, \mathrm{n} \mathrm{is} \mathrm{even}. \mathrm{Then}, \\ \begin{array}{l}\underset{\mathrm{x}\to \infty }{\lim } {\left(2^{1+3+5+.....\frac{\mathrm{n}}{2} \mathrm{terms} }{.2}^{2+4+6+....+\frac{\mathrm{n}}{2} \mathrm{terms}}\right)}^{\frac{1}{\left({\mathrm{n}}^2+1\right)}}\\ =\underset{\mathrm{x}\to \infty }{\lim } {\left(2^{\frac{{\mathrm{n}}^2}{4}}. 3^{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{4}}\right)}^{\frac{1}{\left({\mathrm{n}}^2+1\right)}}\\ =\underset{\mathrm{x}\to \infty }{\lim } 2^{\frac{{\mathrm{n}}^2}{4\left({\mathrm{n}}^2+1\right)}}. 3^{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{4\left({\mathrm{n}}^2+1\right)}}\\ =\underset{\mathrm{x}\to \infty }{\lim } 2 { }^{\underset{\mathrm{x}\to \infty }{\lim } \frac{1}{4\left(1+\frac{1}{{\mathrm{n}}^2}\right)}}. 3 { }^{\underset{\mathrm{x}\to \infty }{\lim } \frac{\left(1+\frac{2}{\mathrm{n}}\right)}{4\left(1+\frac{1}{{\mathrm{n}}^2}\right)}}\\ = 2^{\frac{1}{4}} 3^{\frac{1}{4}}={\left(6\right)}^{\frac{1}{4}}\end{array} \end{gathered}$$