The sum to 10 terms of the series $$\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \ldots$$ is:

We need to find the sum of 10 terms of the series $$\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \ldots$$

To begin,

The general term of the series is:

$$ T_n = \frac{n}{1 + n^2 + n^4} $$

Next,

We need to factorize $$1 + n^2 + n^4$$. This can be done using the identity:

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

To verify: $$(n^2 + n + 1)(n^2 - n + 1) = n^4 - n^3 + n^2 + n^3 - n^2 + n + n^2 - n + 1 = n^4 + n^2 + 1$$ .

So:

$$ T_n = \frac{n}{(n^2 - n + 1)(n^2 + n + 1)} $$

From here,

Notice that $$n^2 + n + 1 - (n^2 - n + 1) = 2n$$. This means $$n = \frac{1}{2}[(n^2 + n + 1) - (n^2 - n + 1)]$$. So:

$$ T_n = \frac{1}{2} \cdot \frac{(n^2 + n + 1) - (n^2 - n + 1)}{(n^2 - n + 1)(n^2 + n + 1)} $$

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

Continuing,

Let $$f(n) = n^2 + n + 1$$. Then:

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

So the general term becomes:

$$ T_n = \frac{1}{2}\left[\frac{1}{f(n-1)} - \frac{1}{f(n)}\right] $$

This is a telescoping series!

Now,

$$ S_{10} = \sum_{n=1}^{10} T_n = \frac{1}{2} \sum_{n=1}^{10} \left[\frac{1}{f(n-1)} - \frac{1}{f(n)}\right] $$

Writing out the terms:

$$ = \frac{1}{2}\left[\left(\frac{1}{f(0)} - \frac{1}{f(1)}\right) + \left(\frac{1}{f(1)} - \frac{1}{f(2)}\right) + \cdots + \left(\frac{1}{f(9)} - \frac{1}{f(10)}\right)\right] $$

Most terms cancel (telescope), leaving only the first and last:

$$ S_{10} = \frac{1}{2}\left[\frac{1}{f(0)} - \frac{1}{f(10)}\right] $$

Moving forward,

$$f(0) = 0^2 + 0 + 1 = 1$$

$$f(10) = 10^2 + 10 + 1 = 100 + 10 + 1 = 111$$

It follows that

$$ S_{10} = \frac{1}{2}\left[1 - \frac{1}{111}\right] = \frac{1}{2} \cdot \frac{111 - 1}{111} = \frac{1}{2} \cdot \frac{110}{111} = \frac{55}{111} $$

The sum to 10 terms is $$\dfrac{55}{111}$$.

The correct answer is Option 2: $$\dfrac{55}{111}$$.