jmc

Let $$f(m)=m^4+\frac{1}{4}=\frac{4m^4+1}{4}.$$We can factor this with a little give and take: $$\begin{array}{rl}f(m)& =\frac{4m^4+1}{4}\\ & =\frac{4m^4+4m^2+1-4m^2}{4}\\ & =\frac{(2m^2+1{)}^2-(2m{)}^2}{4}\\ & =\frac{(2m^2+2m+1)(2m^2-2m+1)}{4}.\end{array}$$Now, let $g(m)=2m^2+2m+1.$ Then $$g(m-1)=2(m-1{)}^2+2(m-1)+1=2m^2-2m+1.$$Hence, $$f(m)=\frac{g(m)g(m-1)}{4}.$$Therefore, $$\begin{array}{rl}\frac{(2^4+\frac{1}{4})(4^4+\frac{1}{4})\cdots [(2n{)}^4+\frac{1}{4}]}{(1^4+\frac{1}{4})(3^4+\frac{1}{4})\cdots [(2n-1{)}^4+\frac{1}{4}]}& =\frac{f(2)f(4)\cdots f(2n)}{f(1)f(3)\cdots f(2n-1)}\\ & =\frac{\frac{g(2)g(1)}{4}\cdot \frac{g(4)g(3)}{4}\cdots \frac{g(2n)g(2n-1)}{4}}{\frac{g(1)g(0)}{4}\cdot \frac{g(3)g(2)}{4}\cdots \frac{g(2n-1)g(2n-2)}{4}}\\ & =\frac{g(2n)}{g(0)}\\ & =2(2n{)}^2+2(2n)+1\\ & =\boxed{8n^2+4n+1}.\end{array}$$