The sum of the first n terms of the series fr | vedclass

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(C) Let the sum be $S_n = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$. The general term $T_k$

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$\frac{n}{2(3n+2)}$

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(C) Let the sum be $S_n = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$. The general term $T_k$ is given by $\frac{1}{(3k-1)(3k+2)}$. We can write $T_k = \frac{1}{3} \left( \frac{3}{(3k-1)(3k+2)} \right) = \frac{1}{3} \left( \frac{(3k+2) - (3k-1)}{(3k-1)(3k+2)} \right) = \frac{1}{3} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$. Now,$S_n = \sum_{k=1}^{n} T_k = \frac{1}{3} \sum_{k=1}^{n} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$. This is a telescoping series: $S_n = \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \ldots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) \right]$. All intermediate terms cancel out,leaving: $S_n = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{3n+2} \right) = \frac{1}{3} \left( \frac{3n+2-2}{2(3n+2)} \right) = \frac{1}{3} \left( \frac{3n}{2(3n+2)} \right) = \frac{n}{2(3n+2)}$.

The sum of the first $n$ terms of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ is

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