frac{1}{2 cdot 5} + frac{1}{5 cdot 8} + frac{ | vedclass

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(C) Let the sum be $S_n = \sum_{k=1}^{n} \frac{1}{(3k-1)(3k+2)}$.We can write the general term as:$\frac{1}{(3k-1)(3k+2)} = \frac{1}{3} \left( \frac{1

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$\frac{n}{6n+4}$

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(C) Let the sum be $S_n = \sum_{k=1}^{n} \frac{1}{(3k-1)(3k+2)}$.We can write the general term as:$\frac{1}{(3k-1)(3k+2)} = \frac{1}{3} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$.Now,summing from $k=1$ to $n$:$S_n = \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \ldots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) \right]$.This is a telescoping series where intermediate terms cancel out:$S_n = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{3n+2} \right)$.$S_n = \frac{1}{3} \left( \frac{3n+2-2}{2(3n+2)} \right) = \frac{1}{3} \left( \frac{3n}{6n+4} \right) = \frac{n}{6n+4}$.

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$ is equal to

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