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 $$\frac{m}{n}$$, where $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to __________.

The general term of the given series is $$T_k=\dfrac{4k}{\,1+4k^{4}\,},\;k\ge 1$$. We need the sum of the first ten terms, $$S_{10}=\displaystyle\sum_{k=1}^{10} T_k$$.

Step 1 – Convert $$T_k$$ into a telescoping form.
We try to express $$T_k$$ as a difference of two simpler rational terms. Write two quadratic expressions

$$A_k=2k^{2}-2k+1,\qquad B_k=2k^{2}+2k+1$$

First compute their product:

$$A_kB_k=(2k^{2}-2k+1)(2k^{2}+2k+1)=4k^{4}+1$$

Now observe

$$\dfrac{1}{A_k}-\dfrac{1}{B_k}= \dfrac{B_k-A_k}{A_kB_k}= \dfrac{4k}{\,4k^{4}+1\,}.$$

Hence

$$T_k=\dfrac{4k}{\,1+4k^{4}\,}= \dfrac{1}{2k^{2}-2k+1}-\dfrac{1}{2k^{2}+2k+1}.$$

Step 2 – Show the series telescopes.
Notice that

$$2k^{2}+2k+1 = 2(k+1)^{2}-2(k+1)+1,$$

so $$\dfrac{1}{2k^{2}+2k+1}= \dfrac{1}{2(k+1)^{2}-2(k+1)+1}.$$ Define

$$a_k=\dfrac{1}{2k^{2}-2k+1}.$$

Then $$T_k = a_k - a_{k+1}.$$

Step 3 – Sum the first ten terms.
Because $$T_k=a_k-a_{k+1},$$ the partial sum is

$$S_{10}= \sum_{k=1}^{10}(a_k-a_{k+1}) =\bigl(a_1-a_2\bigr)+\bigl(a_2-a_3\bigr)+\dots+\bigl(a_{10}-a_{11}\bigr).$$

All intermediate terms cancel, leaving

$$S_{10}=a_1-a_{11}.$$

Step 4 – Evaluate $$a_1$$ and $$a_{11}$$.

$$a_1=\dfrac{1}{2(1)^2-2(1)+1}= \dfrac{1}{2-2+1}=1.$$

$$a_{11}=\dfrac{1}{2(11)^2-2(11)+1}= \dfrac{1}{242-22+1}= \dfrac{1}{221}.$$

Step 5 – Compute $$S_{10}$$.

$$S_{10}=1-\dfrac{1}{221}= \dfrac{221-1}{221}= \dfrac{220}{221}.$$

The fraction $$\dfrac{220}{221}$$ is already in lowest terms because $$221=13\cdot17$$ shares no common factor with $$220=2^{2}\cdot5\cdot11$$.

Thus $$m=220,\;n=221\; \Longrightarrow\; m+n=220+221=441.$$

Hence the required value is $$\boxed{441}$$.