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

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(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.F

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$\frac{4}{25}$

Vedclass · Answer

(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.For $n=16$,the last term is $\frac{1}{(3(16)-1)(3(16)+2)} = \frac{1}{47 \cdot 50}$.The sum $S_{16} = \sum_{n=1}^{16} \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.$S_{16} = \frac{1}{3} \left[ (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{8}) + \ldots + (\frac{1}{47} - \frac{1}{50}) \right]$.$S_{16} = \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{50} \right] = \frac{1}{3} \left[ \frac{25-1}{50} \right] = \frac{1}{3} \left[ \frac{24}{50} \right] = \frac{8}{50} = \frac{4}{25}$.

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ to $16$ terms $=$

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