The sum of the series frac{1}{1 + sqrt{2}} + | vedclass

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(C) The given series is $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$The $n^{th}$ term is $T_n = \f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The given series is $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$The $n^{th}$ term is $T_n = \frac{1}{\sqrt{n} + \sqrt{n+1}}$.By rationalizing the denominator,we get:$T_n = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1} + \sqrt{n})(\sqrt{n+1} - \sqrt{n})} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$.Sum of $15$ terms $S_{15} = \sum_{n=1}^{15} (\sqrt{n+1} - \sqrt{n})$.Expanding the sum:$S_{15} = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{16} - \sqrt{15})$.This is a telescoping series where intermediate terms cancel out:$S_{15} = -\sqrt{1} + \sqrt{16} = -1 + 4 = 3$.Hence,the required sum is $3$.

The sum of the series $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$ up to $15$ terms is

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