Sum of First Ten Terms of Given Series

To find the sum of the first $10$ terms of the given series, we start by identifying the general term $T_k$ of the series:

$T_k = \frac{k}{1 + 4k^4}$

Using Sophie Germain's identity, we can factor the denominator: $1 + 4k^4 = (2k^2)^2 + 4k^2 + 1 - 4k^2 = (2k^2 + 1)^2 - (2k)^2 = (2k^2 - 2k + 1)(2k^2 + 2k + 1)$

Now, observe that the difference between the two factors in the denominator is: $(2k^2 + 2k + 1) - (2k^2 - 2k + 1) = 4k$

Thus, we can rewrite the numerator $k$ as: $k = \frac{1}{4} \left[ (2k^2 + 2k + 1) - (2k^2 - 2k + 1) \right]$

Substituting this back into $T_k$, we get: $T_k = \frac{1}{4} \left[ \frac{(2k^2 + 2k + 1) - (2k^2 - 2k + 1)}{(2k^2 - 2k + 1)(2k^2 + 2k + 1)} \right]$ $T_k = \frac{1}{4} \left[ \frac{1}{2k^2 - 2k + 1} - \frac{1}{2k^2 + 2k + 1} \right]$

Let $V(k) = \frac{1}{2k^2 - 2k + 1}$. Then $V(k+1) = \frac{1}{2(k+1)^2 - 2(k+1) + 1} = \frac{1}{2k^2 + 2k + 1}$.

Therefore, $T_k$ is represented as a telescoping difference: $T_k = \frac{1}{4} [V(k) - V(k+1)]$

The sum of the first $10$ terms, $S_{10}$, is given by: $S_{10} = \sum_{k=1}^{10} T_k = \frac{1}{4} \sum_{k=1}^{10} [V(k) - V(k+1)]$ $S_{10} = \frac{1}{4} [V(1) - V(11)]$

Calculating $V(1)$ and $V(11)$: $V(1) = \frac{1}{2(1)^2 - 2(1) + 1} = 1$ $V(11) = \frac{1}{2(11)^2 - 2(11) + 1} = \frac{1}{2(121) - 22 + 1} = \frac{1}{221}$

Substituting these values into the sum: $S_{10} = \frac{1}{4} \left( 1 - \frac{1}{221} \right) = \frac{1}{4} \left( \frac{220}{221} \right) = \frac{55}{221}$

We are given that $S_{10} = \frac{m}{n}$ with $\text{gcd}(m, n) = 1$. Since $\text{gcd}(55, 221) = 1$, we have: $m = 55 \quad \text{and} \quad n = 221$

Thus, the required value of $m + n$ is: $m + n = 55 + 221 = 276$