JEE Main 2026 — 5 April, Evening Shift — Question 29

The general term is $T_r = \frac{r}{1+4r^4}$. Rewrite denominator: $1+4r^4 = 4r^4+4r^2+1 - 4r^2 = (2r^2+1)^2 - (2r)^2 = (2r^2+2r+1)(2r^2-2r+1)$. Thus $T_r = \frac{r}{(2r^2+2r+1)(2r^2-2r+1)}$. Observe that $\frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1} = \frac{4r}{(2r^2-2r+1)(2r^2+2r+1)}$. Hence $T_r = \frac{1}{4} \left( \frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1} \right)$. Sum from $r=1$ to $10$: $S_{10} = \frac{1}{4} \sum_{r=1}^{10} \left( \frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1} \right)$. This telescopes: $S_{10} = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{2(10)^2+2(10)+1} \right) = \frac{1}{4} \left( 1 - \frac{1}{221} \right) = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$. Thus $m=55, n=221$, $\gcd(55,221)=1$, so $m+n=276$.