The sum of the series frac{1}{2 cdot 5} + fra | vedclass

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Question

(C) The $k$-th term of the series is $T_k = \frac{1}{(3k-1)(3k+2)}$.We can write this as $T_k = \frac{1}{3} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \ri

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

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(C) The $k$-th term of the series is $T_k = \frac{1}{(3k-1)(3k+2)}$.We can write this as $T_k = \frac{1}{3} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$.Summing up to $n$ terms:$S_n = \sum_{k=1}^{n} T_k = \frac{1}{3} \sum_{k=1}^{n} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$.This is a telescoping series:$S_n = \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \dots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) \right]$.$S_n = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{3n+2} \right) = \frac{1}{3} \left( \frac{3n+2-2}{2(3n+2)} \right) = \frac{1}{3} \left( \frac{3n}{6n+4} \right) = \frac{n}{6n+4}$.

The sum of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots$ up to $n$ terms is equal to:

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