JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 27

General term. $\dfrac n{4n^4+1}$, with $4n^4+1=(2n^2-2n+1)(2n^2+2n+1)$, so the term is $\frac14\left[\frac1{2n^2-2n+1}-\frac1{2n^2+2n+1}\right]$.

Telescope to $n=10$: $\frac14\left(1-\frac1{221}\right)=\frac{55}{221}$.

So $m+n=276$.

Answer: 276