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Summation by Sigma Method

The value of ...

Question

The value of $\sum_{n = 1}^{\infty} \frac{1}{(3 n - 2) (3 n + 1)}$ is equal to $\frac{p}{q}$ , where $p$ and $q$ are relatively prime natural numbers. Then the value of $(p^{2} + q^{2})$ is equal to

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Solution

The correct option is C $10$
Given, $S = \infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)} = \frac{1}{3} \infty \sum n = 1 [\frac{1}{3 n - 2} - \frac{1}{3 n + 1}]$
$S = \frac{1}{3} [(1 - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{10}) + . . . .]$
$S = \frac{1}{3} (1 - 0) = \frac{1}{3}$
Comparing with $S = \frac{p}{q} \Rightarrow p = 1 q = 3$
$p^{2} + q^{2} = 1^{2} + (3)^{2} = 10$
$∴$ Option C is correct

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