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:

Step 1: Identify the general term

First, we need to identify the general term of the given series. By observing the pattern in the numerators and denominators, we can express the $k$ -th term as $T_k = \frac{4k}{1 + 4k^4}$. This formula accurately represents each term in the series.

Step 2: Rewrite the denominator using Sophie Germain Identity

The denominator $1 + 4k^4$ can be factored using the Sophie Germain Identity, which states $a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$. Here, we can let $a=1$ and $b=k$. This factorization is crucial for simplifying the term.

Step 3: Decompose the general term using partial fractions

Now, we can decompose the general term $T_k$ into a difference of two fractions. Notice that the numerator $4k$ is the difference between the two factors in the denominator: $(2k^2 + 2k + 1) - (2k^2 - 2k + 1) = 4k$. This allows us to express $T_k$ as a telescoping sum.

Step 4: Define a new function for simplification

Let's define $f(k) = 2k^2 - 2k + 1$. Then, we can observe that $f(k+1) = 2(k+1)^2 - 2(k+1) + 1 = 2(k^2 + 2k + 1) - 2k - 2 + 1 = 2k^2 + 4k + 2 - 2k - 2 + 1 = 2k^2 + 2k + 1$. This means our general term can be written as $T_k = \frac{1}{f(k)} - \frac{1}{f(k+1)}$.

Step 5: Calculate the sum of the first 10 terms

The sum of the first 10 terms is a telescoping series. When we expand the sum, intermediate terms cancel out. We are left with the first term of the first fraction and the last term of the second fraction. We calculate $f(1) = 2(1)^2 - 2(1) + 1 = 1$ and $f(11) = 2(11)^2 - 2(11) + 1 = 2(121) - 22 + 1 = 242 - 22 + 1 = 221$.

Step 6: Compute the final sum and m + n

Substituting the values of $f(1)$ and $f(11)$ into the sum, we get $S_{10} = 1 - \frac{1}{221} = \frac{221 - 1}{221} = \frac{220}{221}$. Here, $m = 220$ and $n = 221$. Since $\text{gcd}(220, 221) = 1$, we can find $m+n = 220 + 221 = 441$.

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