Question

The sum to $n$ term of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots \text { is }$

(A) $\sqrt{2 n+1}$

(B) $\sqrt{2 n+1}-1$

(C) $\frac{1}{2} \sqrt{2 n+1}$

(D) $\frac{1}{2}(\sqrt{2 n+1}-1)$

Answer 1

$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots$

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

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

So, The correct option will be (D).