frac{1}{1 times 5} + frac{1}{5 times 9} + fra | vedclass

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(C) The general term $T_n$ of the series is given by $T_n = \frac{1}{(4n-3)(4n+1)}$.Using partial fractions,we can write $T_n = \frac{1}{4} \left( \fr

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$\frac{n}{4n+1}$

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(C) The general term $T_n$ of the series is given by $T_n = \frac{1}{(4n-3)(4n+1)}$.Using partial fractions,we can write $T_n = \frac{1}{4} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right)$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \frac{1}{4} \sum_{k=1}^{n} \left( \frac{1}{4k-3} - \frac{1}{4k+1} \right)$.Expanding the sum,we get $S_n = \frac{1}{4} \left[ \left( 1 - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{9} \right) + \ldots + \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \right]$.This is a telescoping series,so $S_n = \frac{1}{4} \left( 1 - \frac{1}{4n+1} \right)$.Simplifying,$S_n = \frac{1}{4} \left( \frac{4n+1-1}{4n+1} \right) = \frac{1}{4} \left( \frac{4n}{4n+1} \right) = \frac{n}{4n+1}$.

$\frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \frac{1}{9 \times 13} + \ldots$ up to $n$ terms $=$

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