<s>[INST] If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$. 
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.

The question can be simplified to:

Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.

As $1+n^2+n^4$ forms a GP,
$$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$
But I can't figure out how to solve further. It would be great if someone could help.
 [/INST] HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$
$$\dfrac{2n}{n^4+n^2+1}=\dfrac{(n^2+n+1)-(n^2-n+1)}{(n^2-n+1)(n^2+n+1)}=?$$
If $f(m)=\dfrac1{m^2-m+1}, f(m+1)=?$
See Telescoping series
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