JEE Main 2026 Jan 28 Shift 2, Mathematics Q21: Summation of Series

JEE Main2026Jan 28, Shift 2Mathematics

If $r = 1 \sum 25 (\frac{r}{r ^{4} + r ^{2} + 1}) = \frac{p}{q}$ , where $p$ and $q$ are positive integers such that $g cd (p, q) = 1$ , then $p + q$ is equal to ______.

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Solution

Factor $r^{4} + r^{2} + 1 = (r^{2} + r + 1) (r^{2} - r + 1)$ . So

$\frac{r}{r ^{4} + r ^{2} + 1} = \frac{r}{( r ^{2} - r + 1 ) ( r ^{2} + r + 1 )} = \frac{1}{2} (\frac{1}{r ^{2} - r + 1} - \frac{1}{r ^{2} + r + 1})$ .

This telescopes: $S = \frac{1}{2} r = 1 \sum 25 (\frac{1}{r ^{2} - r + 1} - \frac{1}{r ^{2} + r + 1}) = \frac{1}{2} [\frac{1}{1} - \frac{1}{651}] = \frac{1}{2} \cdot \frac{650}{651} = \frac{325}{651}$ .

(Note: at $r = 25$ , $r^{2} + r + 1 = 651$ .) $g cd (325, 651) = 1$ , so $p + q = 325 + 651 = 976$ .

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JEE Main 2026 Jan 28 Shift 2, Mathematics Q21: Summation of Series

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