A) 0
B) 1
C) 2
D) 4
Correct Answer: C
Solution :
We have, \[\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\] \[+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\] On rationalizing each of the above number separately, we get \[\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\] \[+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}\] \[=-(1-\sqrt{2})-(\sqrt{2}-\sqrt{3})-(\sqrt{3}-\sqrt{4})\] \[-(\sqrt{4}-\sqrt{5})-(\sqrt{5}-\sqrt{6})-(\sqrt{6}-\sqrt{7})-(\sqrt{7}-\sqrt{8})\] \[-(\sqrt{8}-\sqrt{9})\] \[=-1+\sqrt{9}=-1+3=2\]
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