Q49 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
If $\sum_{r=1}^{25}\left(\frac{r}{r^{4}+r^{2}+1}\right)=\frac{p}{q}$, where $p$ and $q$ are positive integers such that $\operatorname{gcd}(p, q)=1$, then $p+q$ is equal to $\_\_\_\_$。
Correct answer: 976
Solution
$S = \sum \dfrac{r}{(r^2 + r + 1)(r^2 - r + 1)}$
$= \dfrac{1}{2} \sum_{r=1}^{25} \left( \dfrac{1}{r^2 - r + 1} - \dfrac{1}{r^2 + r + 1} \right)$
$= \dfrac{1}{2} \left[ \left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \left( \dfrac{1}{3} - \dfrac{1}{7} \right) + \ldots + \left( \dfrac{1}{601} - \dfrac{1}{651} \right) \right]$
$= \dfrac{1}{2} \left[ \dfrac{1}{1} - \dfrac{1}{651} \right]$
$= \dfrac{1}{2} \left[ \dfrac{650}{651} \right] = \dfrac{325}{651}$
$\dfrac{p}{q} = \dfrac{325}{651} \Rightarrow p + q = 976$
$= \dfrac{1}{2} \sum_{r=1}^{25} \left( \dfrac{1}{r^2 - r + 1} - \dfrac{1}{r^2 + r + 1} \right)$
$= \dfrac{1}{2} \left[ \left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \left( \dfrac{1}{3} - \dfrac{1}{7} \right) + \ldots + \left( \dfrac{1}{601} - \dfrac{1}{651} \right) \right]$
$= \dfrac{1}{2} \left[ \dfrac{1}{1} - \dfrac{1}{651} \right]$
$= \dfrac{1}{2} \left[ \dfrac{650}{651} \right] = \dfrac{325}{651}$
$\dfrac{p}{q} = \dfrac{325}{651} \Rightarrow p + q = 976$
Hence, the correct answer is 976.