For positive integers n, if and , then the value of is : [2025]

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Solution

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

1 675

2 1350

3 540

4 135

Ans.
(1)

We have, $a_{n} = \frac{n^{2} + 5 n + 6}{4}$

Also, $S_{n} = \sum_{k = 1}^{n} \frac{1}{a_{k}}$

$= \sum_{k = 1}^{n} \frac{4}{k^{2} + 5 k + 6}$

$= 4 \sum_{k = 1}^{n} \frac{1}{(k + 2) (k + 3)}$

$= 4 \sum_{k = 1}^{n} \frac{1}{k + 2} - \frac{1}{k + 3}$

$= 4 (\frac{1}{3} - \frac{1}{4}) + 4 (\frac{1}{4} - \frac{1}{5}) + . . . + 4 (\frac{1}{n + 2} - \frac{1}{n + 3})$

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

So, $507 S_{2025} = \frac{507 (4) (2025)}{3 [2025 + 3]} = 675$ .

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