If the sum of the first ten terms of the series $$\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \ldots$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are co-prime numbers, then $$m + n$$ is equal to ______

We need to find the sum of the first 10 terms of the series $$\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \ldots$$

First, we check the denominators: $$5, 65, 325, 1025, 2501, \ldots$$

Specifically, for $$n = 1$$ we have $$4(1)^4 + 1 = 5$$ ✓

Similarly, for $$n = 2$$ we get $$4(2)^4 + 1 = 4(16) + 1 = 65$$ ✓

Similarly, for $$n = 3$$ it is $$4(3)^4 + 1 = 4(81) + 1 = 325$$ ✓

Similarly, for $$n = 4$$ it becomes $$4(4)^4 + 1 = 4(256) + 1 = 1025$$ ✓

And for $$n = 5$$ it yields $$4(5)^4 + 1 = 4(625) + 1 = 2501$$ ✓

Therefore, the general term is $$a_n = \frac{n}{4n^4 + 1}$$.

Next, we factor the denominator using the Sophie Germain identity.

The Sophie Germain identity states: $$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$$.

Substituting $$a = 1$$ and $$b = n$$ gives $$4n^4 + 1 = (2n^2 + 2n + 1)(2n^2 - 2n + 1)$$.

For example, when $$n = 1$$: $$(2 + 2 + 1)(2 - 2 + 1) = 5 \times 1 = 5$$ ✓

And when $$n = 2$$: $$(8 + 4 + 1)(8 - 4 + 1) = 13 \times 5 = 65$$ ✓

Then we perform partial fraction decomposition.

Notice that $$(2n^2 + 2n + 1) - (2n^2 - 2n + 1) = 4n$$.

Therefore, $$\frac{n}{(2n^2+2n+1)(2n^2-2n+1)} = \frac{1}{4}\cdot\frac{4n}{(2n^2+2n+1)(2n^2-2n+1)}$$.

$$= \frac{1}{4}\left[\frac{1}{2n^2-2n+1} - \frac{1}{2n^2+2n+1}\right]$$

Next, we identify the telescoping pattern.

Defining $$f(n) = 2n^2 - 2n + 1$$, we have $$f(n+1) = 2(n+1)^2 - 2(n+1) + 1 = 2n^2 + 2n + 1$$.

Hence, $$a_n = \frac{1}{4}\left[\frac{1}{f(n)} - \frac{1}{f(n+1)}\right]$$.

Now, we sum the telescoping series.

Thus, $$S_{10} = \frac{1}{4}\sum_{n=1}^{10}\left[\frac{1}{f(n)} - \frac{1}{f(n+1)}\right] = \frac{1}{4}\left[\frac{1}{f(1)} - \frac{1}{f(11)}\right]$$

To compute the boundary values, note that

$$f(1) = 2(1) - 2(1) + 1 = 1$$

$$f(11) = 2(121) - 2(11) + 1 = 242 - 22 + 1 = 221$$

Substituting these values gives $$S_{10} = \frac{1}{4}\left[1 - \frac{1}{221}\right] = \frac{1}{4}\cdot\frac{220}{221} = \frac{55}{221}$$

Finally, we verify the fraction is in lowest terms and determine $$m + n$$.

$$55 = 5 \times 11$$

$$221 = 13 \times 17$$

Since $$\gcd(55, 221) = 1$$, we have $$m = 55$$ and $$n = 221$$.

$$m + n = 55 + 221 = 276$$

The answer is $$\boxed{276}$$.