11th Class Mathematics Other Series Question Bank Critical Thinking

  • 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) \[n-1\]

    Correct Answer: D

    Solution :

    \[\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+....+\frac{1}{\sqrt{{{n}^{2}}}+\sqrt{{{n}^{2}}-1}}\] 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.


You need to login to perform this action.
You will be redirected in 3 sec spinner