If the sum of the first 20 terms of the series $$\frac{4 \cdot 1}{4 + 3 \cdot 1^2 + 1^4} + \frac{4 \cdot 2}{4 + 3 \cdot 2^2 + 2^4} + \frac{4 \cdot 3}{4 + 3 \cdot 3^2 + 3^4} + \frac{4 \cdot 4}{4 + 3 \cdot 4^2 + 4^4} + \ldots$$ is $$\frac{m}{n}$$, where m and n are coprime, then $$m + n$$ is equal to :

Let the general term of the given series be $$T_r$$:

$$T_r = \frac{4r}{4 + 3r^2 + r^4}$$

The denominator can be factorized by completing the square:

$$r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$$

Using the difference of squares identity $$A^2 - B^2 = (A - B)(A + B)$$:

$$r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$$

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Now, we rewrite the numerator $$4r$$ in terms of these two factors:

$$4r = 2 \left[ (r^2 + r + 2) - (r^2 - r + 2) \right]$$

Substituting this back into the expression for $$T_r$$:

$$T_r = \frac{2 \left[ (r^2 + r + 2) - (r^2 - r + 2) \right]}{(r^2 - r + 2)(r^2 + r + 2)}$$

$$T_r = \frac{2}{r^2 - r + 2} - \frac{2}{r^2 + r + 2}$$

This can be written in telescoping form as $$T_r = V_r - V_{r+1}$$, where $$V_r = \frac{2}{r^2 - r + 2}$$.

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The sum of the first 20 terms is given by:

$$S_{20} = \sum_{r=1}^{20} T_r = \sum_{r=1}^{20} (V_r - V_{r+1})$$

$$S_{20} = (V_1 - V_2) + (V_2 - V_3) + \dots + (V_{20} - V_{21})$$

$$S_{20} = V_1 - V_{21}$$

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Calculating the values of $$V_1$$ and $$V_{21}$$:

$$V_1 = \frac{2}{1^2 - 1 + 2} = \frac{2}{2} = 1$$

$$V_{21} = \frac{2}{21^2 - 21 + 2} = \frac{2}{441 - 21 + 2} = \frac{2}{422} = \frac{1}{211}$$

Substituting these back into the sum:

$$S_{20} = 1 - \frac{1}{211} = \frac{210}{211}$$

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Comparing this with $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime:

$$m = 210, \quad n = 211$$

The required value is:

$$m + n = 210 + 211 = 421$$

Therefore, the value of $$m + n$$ is equal to 421.