JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Sequences and Series

Factor the denominator as $k^4+k^2+1=(k^2+k+1)(k^2-k+1)$, and note that $k^2+k+1$ is just $k^2-k+1$ with $k$ replaced by $k+1$. Partial fractions then give $\frac{k}{k^4+k^2+1}=\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)$, a difference between a term and its successor, so the sum telescopes. With $k$ running from $1$ to $10$ only the first lower piece and the last upper piece survive, giving $\frac{1}{2}\left(1-\frac{1}{111}\right)=\frac{1}{2}\cdot \frac{110}{111}=\frac{55}{111}$. Since $55$ and $111$ have no common factor, $m=55$ and $n=111$, so $m+n=166$.