The numerators are $21=5^2-2^2$, $273=17^2-4^2$, $1333=37^2-6^2$, $4161=65^2-8^2$, $10101=101^2-10^2$, and $20881=145^2-12^2$. A little work with finite differences shows that the sequence $5,17,37,65,101,145$ is given by $4n^2+1$, so it appears that your $n$-th term is $\frac{(4n^2+1)^2-(2n)^2}{8n^2}=\frac{(4n^2-2n+1)(4n^2+2n+1)}{8n^2}\;,$ or, if you prefer, $\frac{16n^4+4n^2+1}{8n^2}\;.$
Added: In the second sequence there may be a typo in the third term: it may have been supposed to be $2353/98$. With that emendation the numerator of the $n$-th term would be $2n(n-1)d_n+1$, where $d_n$ is the denominator of the $n$-th term. Of course that still leaves the denominators a bit of a puzzle: $d_n=2(2n-1)^2$ would work except for the $40$, which ought by that rule to be $50$. Before spending more time on it, Iād want to be sure that the sequence has been given correctly.
Added2: The corrected sequence, with $601/50$ as third term, fits the pattern that I suspected: the $n$-th term is $\frac{4n(n-1)(2n-1)^2+1}{2(2n-1)^2}=2n(n-1)+\frac1{2(2n-1)^2}\;.$