For any n in N,frac{1}{2 cdot 5} + frac{1}{5 | vedclass

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(A) Let $S_n = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)}$.The general term is $T_n = \frac{1}{(3n-1)(3n+2)}$.Using p

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

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(A) Let $S_n = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)}$.The general term is $T_n = \frac{1}{(3n-1)(3n+2)}$.Using partial fractions,we can write $T_n = \frac{1}{3} \left[ \frac{1}{3n-1} - \frac{1}{3n+2} \right]$.Summing from $n=1$ to $n$,we get a telescoping series:$S_n = \sum_{k=1}^{n} T_k = \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \ldots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) \right]$.All intermediate terms cancel out:$S_n = \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{3n+2} \right]$.Simplifying the expression:$S_n = \frac{1}{3} \left[ \frac{(3n+2) - 2}{2(3n+2)} \right] = \frac{1}{3} \left[ \frac{3n}{2(3n+2)} \right] = \frac{n}{2(3n+2)} = \frac{n}{6n+4}$.

For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

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