1yes ans is d
\frac{n}{1+n^2+n^4}=\frac{1}{2}\left( \frac{1}{1-n+n^2}-\frac{1}{1+n+n^2}\right)
\sum_{n=1}^{n}{\frac{n}{1+n^2+n^4}}=\frac{1}{2}\left( 1-\frac{1}{1+n+n^2}\right)
all terms are canceled in summation except first and last
so for S(10) & S(20) put n =10&20 in summation
u get S(10) = 55/111 & S(20) = 210/421