If $\sum_{\mathrm{r}=1}^{25}\left(\frac{\mathrm{r}}{\mathrm{r}^{4}+\mathrm{r}^{2}+1}\right)=\frac{\mathrm{p}}{\mathrm{q}}$, where p and q are positive integers such that $\operatorname{gcd}(p, q)=1$, then $p+q$ is equal to ______。

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Sequences and Series

\sum_{r = 1}^{25} (\frac{r}{r ^{4} + r ^{2} + 1}) = \frac{p}{q}

, where p and q are positive integers such that

gcd (p, q) = 1

, then

p + q

is equal to ______。

Answer: 976

Factor

r^{4} + r^{2} + 1 = (r^{2} - r + 1) (r^{2} + r + 1)

. Since

(r^{2} + r + 1) - (r^{2} - r + 1) = 2 r

\frac{r}{( r ^{2} - r + 1 ) ( r ^{2} + r + 1 )} = \frac{1}{2} (\frac{1}{r ^{2} - r + 1} - \frac{1}{r ^{2} + r + 1}) .

Writing

g (r) = \frac{1}{r ^{2} - r + 1}

and noting

r^{2} + r + 1 = (r + 1)^{2} - (r + 1) + 1 = \frac{1}{g ( r + 1 )}

, each term is

\frac{1}{2} (g (r) - g (r + 1))

, so the sum telescopes:

r = 1 \sum 25 \frac{r}{r ^{4} + r ^{2} + 1} = \frac{1}{2} (g (1) - g (26)) = \frac{1}{2} (1 - \frac{1}{651}) = \frac{325}{651} .

Here

325 = 5^{2} \cdot 13

and

651 = 3 \cdot 7 \cdot 31

are coprime, so

p = 325, q = 651

and

p + q = 976

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer