The sum of n terms of the series frac{1}{1 + | vedclass

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(D) The $k$-th term of the series is $a_k = \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}}$.Rationalizing the denominator,we get $a_k = \frac{\sqrt{2k+1} - \sqrt

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$\frac{1}{2}(\sqrt{2n + 1} - 1)$

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(D) The $k$-th term of the series is $a_k = \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}}$.Rationalizing the denominator,we get $a_k = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{(2k+1) - (2k-1)} = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} a_k = \frac{1}{2} \sum_{k=1}^{n} (\sqrt{2k+1} - \sqrt{2k-1})$.Expanding the sum: $S_n = \frac{1}{2} [(\sqrt{3} - 1) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \dots + (\sqrt{2n+1} - \sqrt{2n-1})]$.This is a telescoping series where intermediate terms cancel out,leaving $S_n = \frac{1}{2} (\sqrt{2n+1} - 1)$.

The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

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