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 :-

The given series is $$\displaystyle S_{10}= \sum_{k=1}^{10} \frac{k}{1+k^{2}+k^{4}}$$.

First factor the denominator. Using the identity $$(x^{2}+x+1)(x^{2}-x+1)=x^{4}+x^{2}+1$$, put $$x=k$$ to get $$1+k^{2}+k^{4}=(k^{2}+k+1)(k^{2}-k+1)$$.

Hence each term becomes
$$\frac{k}{(k^{2}-k+1)(k^{2}+k+1)}.$$

We now decompose this fraction. Assume
$$\frac{k}{(k^{2}-k+1)(k^{2}+k+1)} =\frac{A}{k^{2}-k+1}+\frac{B}{k^{2}+k+1}.$$

Cross-multiplying and equating coefficients:

$$A(k^{2}+k+1)+B(k^{2}-k+1)=k.$$

Comparing powers of $$k$$:
Coefficient of $$k^{2}$$: $$A+B=0$$,
Coefficient of $$k$$ : $$A-B=1$$,
Constant term : $$A+B=0$$.

Solving, $$A=\tfrac12$$ and $$B=-\tfrac12$$. Therefore

$$\frac{k}{(k^{2}-k+1)(k^{2}+k+1)} =\frac12\left[\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1}\right].$$

Define $$T_k=\frac12\left[\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1}\right].$$ Notice that $$k^{2}+k+1=(k+1)^{2}-(k+1)+1,$$ so the second fraction of $$T_k$$ equals the first fraction of $$T_{k+1}$$. This gives a telescoping sum.

Sum of the first $$n$$ terms:
$$\sum_{k=1}^{n} T_k =\frac12\!\left[\frac{1}{1^{2}-1+1}-\frac{1}{(n+1)^{2}-(n+1)+1}\right].$$

For $$n=10$$:
$$1^{2}-1+1 = 1,$$ so the first term is $$1$$.
$$(10+1)^{2}-(10+1)+1 = 11^{2}-11+1 = 121-11+1 = 111,$$ so the last term is $$\frac{1}{111}$$.

Therefore
$$S_{10}=\frac12\left[1-\frac1{111}\right] =\frac12\left[\frac{110}{111}\right] =\frac{55}{111}.$$

Thus the required sum is $$\displaystyle \frac{55}{111}$$.

The correct option is Option B.