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

The sum of nn terms of the series 11+3+13+5+15+7+...\dfrac{1}{{1 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 5 }} + \dfrac{1}{{\sqrt 5 + \sqrt 7 }} + ...is
A) 2n+1\sqrt {2n + 1}
B) 122n+1\dfrac{1}{2}\sqrt {2n + 1}
C) 2n−1\sqrt {2n - 1}
D)12(2n+1−1)\dfrac{1}{2}(\sqrt {2n + 1} - 1)

Explanation

Solution

This is a classic example of the telescoping series. We first will have to find the nnth term and then proceed with the calculation by rationalizing the denominators.

Complete step-by-step answer:
By the series we can observe that the two terms in the denominator of the fractions are odd consecutive numbers. By this knowledge we can calculate the terms in the denominator of the nnth term. So the first term in the denominator of the nnth term would be square root of 2n−12n - 1 and the next term would be square root of 2n+12n + 1. Thus, the series can be rewritten as
11+3+13+5+15+7+...+12n−1+2n+1\dfrac{1}{{1 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 5 }} + \dfrac{1}{{\sqrt 5 + \sqrt 7 }} + ... + \dfrac{1}{{\sqrt {2n - 1} + \sqrt {2n + 1} }}
Now rationalizing the denominators, we get

=(1−3)(1+3)(1−3)+(3−5)(3+5)(3−5)+(5−7)(5+7)(5−7)+...+(2n−1−2n+1)(2n−1+2n+1)(2n−1−2n+1)  =(1−3)−2+(3−5)−2+(5−7)−2+...+(2n−1−2n+1)−2  =1−2n+1−2  =2n+1−12  = \dfrac{{(1 - \sqrt 3 )}}{{(1 + \sqrt 3 )(1 - \sqrt 3 )}} + \dfrac{{(\sqrt 3 - \sqrt 5 )}}{{(\sqrt 3 + \sqrt 5 )(\sqrt 3 - \sqrt 5 )}} + \dfrac{{(\sqrt 5 - \sqrt 7 )}}{{(\sqrt 5 + \sqrt 7 )(\sqrt 5 - \sqrt 7 )}} + ... + \dfrac{{(\sqrt {2n - 1} - \sqrt {2n + 1} )}}{{(\sqrt {2n - 1} + \sqrt {2n + 1} )(\sqrt {2n - 1} - \sqrt {2n + 1} )}} \\\ \\\ = \dfrac{{(1 - \sqrt 3 )}}{{ - 2}} + \dfrac{{(\sqrt 3 - \sqrt 5 )}}{{ - 2}} + \dfrac{{(\sqrt 5 - \sqrt 7 )}}{{ - 2}} + ... + \dfrac{{(\sqrt {2n - 1} - \sqrt {2n + 1} )}}{{ - 2}} \\\ \\\ = \dfrac{{1 - \sqrt {2n + 1} }}{{ - 2}} \\\ \\\ = \dfrac{{\sqrt {2n + 1} - 1}}{2} \\\

Thus, we can see that the correct option is option D.

Note: This problem may seem hard to the eye but this is not the case. If we know the approach to the problem, we can easily solve it. In most of the cases the series gets resolved by rationalizing the terms and then except the first and the last terms, most of the terms get cancelled out. These types of problems can only be solved through ample amounts of practice, because then we get to see the variety of sums in this domain. The telescoping series can be that of a trigonometric function and if not properly dealt with, it can lead to a lot of confusion. So the trick here is to figure the different ways in which the problem can be solved.