If the sum of the first 10 terms of the series (1)/(1 + 1^4 × 4) +...

The general term of the series is $T_r = \dfrac{r}{1 + 4r^4}$.

Factorizing the denominator:

$1 + 4r^4 = (1 + 2r^2)^2 - 4r^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$

Thus, $T_r = \dfrac{r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$

Multiplying and dividing by $4$:

$T_r = \dfrac{1}{4} \left[ \dfrac{4r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)} \right]$

$T_r = \dfrac{1}{4} \left[ \dfrac{1}{2r^2 - 2r + 1} - \dfrac{1}{2r^2 + 2r + 1} \right]$

The sum of the first $10$ terms is:

$S_{10} = \sum_{r=1}^{10} T_r = \dfrac{1}{4} \sum_{r=1}^{10} \left( \dfrac{1}{2r^2 - 2r + 1} - \dfrac{1}{2(r+1)^2 - 2(r+1) + 1} \right)$

This is a telescoping series, so all intermediate terms cancel out:

$S_{10} = \dfrac{1}{4} \left[ \left( \dfrac{1}{1} - \dfrac{1}{5} \right) + \left( \dfrac{1}{5} - \dfrac{1}{13} \right) + \ldots + \left( \dfrac{1}{2(10)^2 - 2(10) + 1} - \dfrac{1}{2(10)^2 + 2(10) + 1} \right) \right]$

$S_{10} = \dfrac{1}{4} \left[ 1 - \dfrac{1}{221} \right] = \dfrac{1}{4} \left( \dfrac{220}{221} \right) = \dfrac{55}{221}$

Given $S_{10} = \dfrac{m}{n}$ and $\gcd(m, n) = 1$.

Here, $m = 55$ and $n = 221$. Since $55 = 5 \times 11$ and $221 = 13 \times 17$, $\gcd(55, 221) = 1$.

Therefore, $m + n = 55 + 221 = 276$.

Answer: $276$