frac{1}{3 cdot 5} + frac{1}{5 cdot 7} + frac{ | vedclass

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(D) The general term $T_n$ of the series is given by $T_n = \frac{1}{(2n+1)(2n+3)}$.We can write this as $T_n = \frac{1}{2} \left( \frac{1}{2n+1} - \f

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$\frac{8}{51}$

Vedclass · Answer

(D) The general term $T_n$ of the series is given by $T_n = \frac{1}{(2n+1)(2n+3)}$.We can write this as $T_n = \frac{1}{2} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right)$.The sum of $n$ terms $S_n = \sum_{k=1}^{n} T_k = \frac{1}{2} \sum_{k=1}^{n} \left( \frac{1}{2k+1} - \frac{1}{2k+3} \right)$.This is a telescoping series: $S_n = \frac{1}{2} \left[ (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \ldots + (\frac{1}{2n+1} - \frac{1}{2n+3}) \right]$.$S_n = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{2n+3} \right)$.For $n = 24$,$S_{24} = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{2(24)+3} \right) = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{51} \right)$.$S_{24} = \frac{1}{2} \left( \frac{17-1}{51} \right) = \frac{1}{2} \cdot \frac{16}{51} = \frac{8}{51}$.

$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots$ to $24$ terms $=$

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