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.
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