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

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(D) The given series is $S = \sum_{k=1}^{n^2-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$.To solve this,we rationalize each term by multiplying the numerator a

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$n - 1$

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(D) The given series is $S = \sum_{k=1}^{n^2-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$.To solve this,we rationalize each term by multiplying the numerator and denominator by the conjugate $(\sqrt{k+1} - \sqrt{k})$:$\frac{1}{\sqrt{k+1} + \sqrt{k}} 	imes \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k}$.Applying this to the series:$S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + ... + (\sqrt{n^2} - \sqrt{n^2 - 1})$.This is a telescoping series where intermediate terms cancel out:$S = -\sqrt{1} + \sqrt{n^2} = -1 + n = n - 1$.

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

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