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(C) The $n$-th term of the series is $T_r = \frac{1}{(2r)(2r+2)} = \frac{1}{4r(r+1)}$.Using partial fractions,$T_r = \frac{1}{4} \left( \frac{1}{r} -

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$1$

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(C) The $n$-th term of the series is $T_r = \frac{1}{(2r)(2r+2)} = \frac{1}{4r(r+1)}$.Using partial fractions,$T_r = \frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} \right)$.The sum of $n$ terms is $S_n = \sum_{r=1}^{n} T_r = \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{r} - \frac{1}{r+1} \right)$.This is a telescoping series: $S_n = \frac{1}{4} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right]$.$S_n = \frac{1}{4} \left( 1 - \frac{1}{n+1} \right) = \frac{1}{4} \left( \frac{n+1-1}{n+1} \right) = \frac{n}{4(n+1)}$.Comparing this with the given expression $\frac{kn}{4(n+1)}$,we get $k = 1$.

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

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