A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
Given series is \[\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+.....\] \[{{\text{n}}^{\text{th}}}\,\text{term}\,\text{=}\frac{1}{\sqrt{n}+\sqrt{n+1}}\] \[\therefore \]\[\text{1}{{\text{5}}^{\text{th}}}\,\text{term}\,\text{=}\frac{1}{\sqrt{15}+\sqrt{16}}\] Thus, given series upto 15 terms is \[\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+.....+\frac{1}{\sqrt{15}+\sqrt{16}}\]This can be re-written as \[\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+.......+\frac{\sqrt{15}-\sqrt{16}}{-1}\] (By rationalization) \[=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+....-\sqrt{14}+\sqrt{15}\]\[-\sqrt{15}+\sqrt{16}\] \[=-1+\sqrt{16}=-1+4=3\] Hence, the required sum = 3
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