If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1+4 \cdot 1^4}+\frac{4 \cdot 2}{1+4 \cdot 2^4}+\frac{4 \cdot 3}{1+4 \cdot 3^4}+\ldots .$. is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to _______________

<p>$$\begin{aligned} & \frac{4.1}{1+4.1^4}+\frac{4.2}{1+4.2^4}+\frac{4.3}{1+4.3^4}+\ldots . \\ & T_r=\frac{4 r}{1+4 r^4}=\frac{4 r}{4 r^4+4 r^2+1-4 r^2} \\ & =\frac{4 r}{\left(2 r^2+1\right)^2-(2 r)^2} \\ & T_r=\frac{4 r}{\left(2 r^2-2 r+1\right)\left(2 r^2+2 r+1\right)} \end{aligned}$$</p> <p>$$\begin{aligned} & T_r=\frac{\left(2 r^2+2 r+1\right)-\left(2 r^2-2 r+1\right)}{\left(2 r^2-2 r+1\right)\left(2 r^2+2 r+1\right)} \\ & T_r=\left(\frac{1}{r^2+(r-1)^2}-\frac{1}{r^2+(r+1)^2}\right) \\ & \sum_{r=1}^{10} T_r=\left(\frac{1}{0^2+1^2}-\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2}-\frac{1}{2^2+3^2}+\ldots .\right. \\ & \frac{1}{9^2+10^2}-\frac{1}{10^2+11^2} \\ & =1-\frac{1}{221} \\ & =\frac{220}{221} \\ & \therefore \quad m+n=220+221 \\ & \quad=441 \end{aligned}$$</p>