JEE Main 2025 Jan 28 Shift 2, Mathematics Q13: Summation of Series

JEE Main2025Jan 28, Shift 2Mathematics

For positive integers $n$ , if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = k = 1 \sum n (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is :

Solution

From $4 a_{n} = n^{2} + 5 n + 6 = (n + 2) (n + 3)$ , $\frac{1}{a _{n}} = \frac{4}{( n + 2 ) ( n + 3 )} = 4 (\frac{1}{n + 2} - \frac{1}{n + 3})$ .

The sum telescopes:

$S_{n} = 4 (\frac{1}{3} - \frac{1}{n + 3}) = \frac{4 n}{3 ( n + 3 )}$ .

For $n = 2025$ : $S_{2025} = \frac{4 \cdot 2025}{3 \cdot 2028}$ , so

$507 S_{2025} = \frac{507 \cdot 4 \cdot 2025}{3 \cdot 2028} = \frac{507 \cdot 4 \cdot 2025}{6084} = 675$ .

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