The sum of the series 1 1 − 3 ⋅ 1 2 + 1 4 + 2 1 − 3 ⋅ 2 2 + 2 4 + 3 1 − 3 ⋅ 3 2 + 3 4 + . ... up to 10 terms…

General term of the given series is given by,

$T_n=\frac{n}{1-3n^2+n^4}$

$\Rightarrow T_n=\frac{n}{\left(n^4-2n^2+1\right)-n^2}$

$\Rightarrow T_n=\frac{n}{{\left(n^2-1\right)}^2-n^2}$

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

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

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

$\Rightarrow \sum _{n=1}^{10}{T}_n=\frac{1}{2}\left[\left(\frac{1}{-1}-\frac{1}{1}\right)+\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{11}\right)+...+\left(\frac{1}{89}-\frac{1}{109}\right)\right]$

$\Rightarrow \sum _{n=1}^{10}{T}_n=\frac{1}{2}\left[-1-\frac{1}{109}\right]$

$\Rightarrow \sum _{n=1}^{10}{T}_n=\frac{-55}{109}$