• question_answer The sum of the series $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{{{n}^{2}}-1}+\sqrt{{{n}^{2}}}}$equals                                          [AMU 2002] A) $\frac{(2n+1)}{\sqrt{n}}$ B) $\frac{\sqrt{n}+1}{\sqrt{n}+\sqrt{n-1}}$ C) $\frac{(n+\sqrt{{{n}^{2}}-1})}{2\sqrt{n}}$ D) $n-1$

$\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+....+\frac{1}{\sqrt{{{n}^{2}}}+\sqrt{{{n}^{2}}-1}}$ Rationalization of ${{D}^{r}}$ $\therefore S=(\sqrt{2}-\sqrt{1})+\left( \sqrt{3}-\sqrt{2} \right)+...+\left( \sqrt{{{n}^{2}}}-\sqrt{{{n}^{2}}-1} \right)$    S = n - 1.