JEE Main 2026 — 28 January, Evening Shift — Question 21

$S=\sum \frac{r}{\left(r^{2}+r+1\right)\left(r^{2}-r+1\right)}$

$$\begin{aligned} & =\frac{1}{2} \sum_{\mathrm{r}=1}^{25}\left(\frac{1}{\mathrm{r}^{2}-\mathrm{r}+1}-\frac{1}{\mathrm{r}^{2}+\mathrm{r}+1}\right) & \end{aligned}$$

$=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\ldots+\left(\frac{1}{601}-\frac{1}{651}\right)\right]$

$=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{651}\right]$ $=\frac{1}{2}\left[\frac{650}{651}\right]=\frac{325}{651}$

$\frac{\mathrm{p}}{\mathrm{q}}=\frac{325}{651}$

$\Rightarrow \mathrm{p}+\mathrm{q}=976$