JEE Main 2026 — Mathematics Algebra

The general term of the series is $T_r = \dfrac{r}{1 + 4r^4}$Tr​=1+4r4r​.

Factorizing the denominator:

$$1 + 4r^4 = (1 + 2r^2)^2 - 4r^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$$1+4r4=(1+2r2)2−4r2=(2r2−2r+1)(2r2+2r+1)

Thus, $T_r = \dfrac{r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$Tr​=(2r2−2r+1)(2r2+2r+1)r​

Multiplying and dividing by $4$4:

$$T_r = \dfrac{1}{4} \left[ \dfrac{4r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)} \right]$$Tr​=41​[(2r2−2r+1)(2r2+2r+1)4r​]

$$T_r = \dfrac{1}{4} \left[ \dfrac{1}{2r^2 - 2r + 1} - \dfrac{1}{2r^2 + 2r + 1} \right]$$Tr​=41​[2r2−2r+11​−2r2+2r+11​]

The sum of the first $10$10 terms is:

$$S_{10} = \sum_{r=1}^{10} T_r = \dfrac{1}{4} \sum_{r=1}^{10} \left( \dfrac{1}{2r^2 - 2r + 1} - \dfrac{1}{2(r+1)^2 - 2(r+1) + 1} \right)$$S10​=r=1∑10​Tr​=41​r=1∑10​(2r2−2r+11​−2(r+1)2−2(r+1)+11​)

This is a telescoping series, so all intermediate terms cancel out:

$$S_{10} = \dfrac{1}{4} \left[ \left( \dfrac{1}{1} - \dfrac{1}{5} \right) + \left( \dfrac{1}{5} - \dfrac{1}{13} \right) + \ldots + \left( \dfrac{1}{2(10)^2 - 2(10) + 1} - \dfrac{1}{2(10)^2 + 2(10) + 1} \right) \right]$$S10​=41​[(11​−51​)+(51​−131​)+…+(2(10)2−2(10)+11​−2(10)2+2(10)+11​)]

$$S_{10} = \dfrac{1}{4} \left[ 1 - \dfrac{1}{221} \right] = \dfrac{1}{4} \left( \dfrac{220}{221} \right) = \dfrac{55}{221}$$S10​=41​[1−2211​]=41​(221220​)=22155​

Given $S_{10} = \dfrac{m}{n}$S10​=nm​ and $\gcd(m, n) = 1$gcd(m,n)=1.

Here, $m = 55$m=55 and $n = 221$n=221. Since $55 = 5 \times 11$55=5×11 and $221 = 13 \times 17$221=13×17, $\gcd(55, 221) = 1$gcd(55,221)=1.

Therefore, $m + n = 55 + 221 = 276$m+n=55+221=276.

Answer: $276$276