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Question: The sum of *n* terms of the series \(\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{...

The sum of n terms of the series 11+3+13+5+15+7+.......\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + ....... is

A

2n+1\sqrt{2n + 1}

B

122n+1\frac{1}{2}\sqrt{2n + 1}

C

2n−1\sqrt{2n - 1}

D

12(2n+1−1)\frac{1}{2}(\sqrt{2n + 1} - 1)

Answer

12(2n+1−1)\frac{1}{2}(\sqrt{2n + 1} - 1)

Explanation

Solution

Sn=11+3+13+5+15+7+......+12n−1+2n+1=3−1(3−1)(3+1)+5−32+7−52+.....+2n+1−2n−12=12[3−1+5−3+7−5+.....+(2n+1−2n−1)]=12[2n+1−1]S_{n} = \frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + ...... + \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}} = \frac{\sqrt{3} - 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + ..... + \frac{\sqrt{2n + 1} - \sqrt{2n - 1}}{2} = \frac{1}{2}\lbrack\sqrt{3} - 1 + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + ..... + (\sqrt{2n + 1} - \sqrt{2n - 1})\rbrack = \frac{1}{2}\lbrack\sqrt{2n + 1} - 1\rbrack