Question

The sum to n terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...$

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

Answer 1

The correct option is D

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

$Let{T}_n\text{be the nth term of the given series.}Thus,wehave:T_n=\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=\frac{\sqrt{2n+1}-\sqrt{2n-1}}{2}{S}_n={\sum }_{k=1}^n{T}_k={\sum }_{k=1}^n\left(\frac{\sqrt{2k+1}-\sqrt{2k-1}}{2}\right)\frac{1}{2}{\sum }_{k=1}^n(\sqrt{2k+1}-\sqrt{2k-1})=\frac{1}{2}[(\sqrt{3}-\sqrt{1}+)+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+...+(\sqrt{2n+1}-\sqrt{2n-1})]=\frac{1}{2}\left\{\right(-1)+\sqrt{2n+1}\}=\frac{1}{2}\{\sqrt{2n+1}-1\}$