Question

$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+......$ upto n terms. is equal to

• A. $\sqrt{2n+1}$
• B. $\frac{1}{\sqrt{2n+1}+1}$
• C. $\frac{n}{\sqrt{2n+1}+1}$
• D. $\frac{n}{\sqrt{2n+1}-1}$

Answer 1

The correct option is C $\frac{n}{\sqrt{2n+1}+1}$

$T_n=\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=\frac{-\sqrt{2n-1}+\sqrt{2n+1}}{2}$

$S_n=\sum T_n$

$=\frac{1}{2}\left[\right(\sqrt{3}-1)+(\sqrt{5}-\sqrt{3})+(\sqrt{7}+\sqrt{5})+....+(\sqrt{2n+1})-\sqrt{2n-1}]$

$=\frac{1}{2}[\sqrt{2n+1}-1]=\frac{n}{\sqrt{2n+1}+1}$