If frac{1}{2 times 4} + frac{1}{4 times 6} + | vedclass

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(A) The given series is $S_n = \sum_{r=1}^{n} \frac{1}{(2r)(2r+2)}$.We can write the general term as $\frac{1}{4} 	imes \frac{1}{r(r+1)} = \frac{1}{4

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$\frac{1}{4}$

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(A) The given series is $S_n = \sum_{r=1}^{n} \frac{1}{(2r)(2r+2)}$.We can write the general term as $\frac{1}{4} 	imes \frac{1}{r(r+1)} = \frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} ight)$.Summing this from $r=1$ to $n$:$S_n = \frac{1}{4} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight]$.This is a telescoping series,so $S_n = \frac{1}{4} \left( 1 - \frac{1}{n+1} ight)$.$S_n = \frac{1}{4} \left( \frac{n+1-1}{n+1} ight) = \frac{n}{4(n+1)}$.Comparing this with $\frac{k n}{n+1}$,we get $k = \frac{1}{4}$.

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to

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