I would like one fourth, please.

First, learn the important factorization. $\begin{aligned} &4a^4+1 \\ =&4a^4+4a^2+1-4a^2 \\ =&(2a^2+1)^2-(2a)^2 \\ =&(2a^2-2a+1)(2a^2+2a+1) \\ =&((a-1)^2+a^2)(a^2+(a+1)^2) \\ \end{aligned}$ Then, calculate it. $\begin{aligned} &\frac{(40^4 + \frac{1}{4})(38^4 + \frac{1}{4}) … (4^4 + \frac{1}{4})(2^4 + \frac{1}{4})} {(39^4 + \frac{1}{4})(37^4 + \frac{1}{4}) …(3^4 + \frac{1}{4})(1^4 + \frac{1}{4})} \\ =&\frac{\prod_{k=1}^{20} [(2k)^4+\frac{1}{4}]}{\prod_{k=1}^{20} [(2k-1)^4+\frac{1}{4}]} \\ =&\frac{\prod_{k=1}^{20} [4(2k)^4+1]}{\prod_{k=1}^{20} [4(2k-1)^4+1]} \\ =&\frac{\prod_{k=1}^{20} {[(2k-1)^2+(2k)^2][(2k)^2+(2k+1)^2]} }{\prod_{k=1}^{20} {[(2k-2)^2+(2k-1)^2][(2k-1)^2+(2k)^2]}} \\ =&\frac{\prod_{k=1}^{20} [(2k)^2+(2k+1)^2]} {\prod_{k=1}^{20} [(2k-2)^2+(2k-1)^2]} \\ =&\frac{(2^2+3^2)(4^2+5^2)…(38^2+39^2)(40^2+41^2)} {(0^2+1^2)(2^2+3^2)…(38^2+39^2)} \\ =&40^2+41^2 \\ =&\large 3281 \end{aligned}$