The sum of 10 terms of the series

$${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$$ is

$$ \begin{aligned} & T_n=\frac{n}{1+n^2+n^4} \\\\ & =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\\\ & =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\ & \Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left[\frac{1}{\left(\mathrm{n}^2-\mathrm{n}+1\right)}-\frac{1}{\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\ & \mathrm{S}_{\mathrm{n}}=\sum_{n=1}^{10} T_n \\\\ & =\frac{1}{2} \sum\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right) \\\\ & =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)\right. \\\\ & \left.\dots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\left(\frac{1}{91}-\frac{1}{111}\right)\right] \end{aligned} $$ <br/><br/>$\therefore$ $ S=\frac{1}{2}\left(1-\frac{1}{111}\right)=\frac{55}{111} $