What is the value of the following sum?
$\frac{1}{(\sqrt{2}+\sqrt{1})} + \frac{1}{(\sqrt{3}+\sqrt{2})} + \frac{1}{(\sqrt{4}+\sqrt{3})} + … + \frac{1}{(\sqrt{100}+\sqrt{99})}$
9
8
7
6
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2024
$\frac{1}{\sqrt{k+1}+\sqrt{k}} \times \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}} = \frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1})^2 – (\sqrt{k})^2} = \frac{\sqrt{k+1}-\sqrt{k}}{(k+1) – k} = \sqrt{k+1}-\sqrt{k}$.
$(\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4}-\sqrt{3}) + \dots + (\sqrt{100}-\sqrt{99})$
The intermediate terms cancel out: $(-\sqrt{1} + \sqrt{2}) + (-\sqrt{2} + \sqrt{3}) + (-\sqrt{3} + \sqrt{4}) + \dots + (-\sqrt{99} + \sqrt{100})$.
The sum simplifies to $\sqrt{100} – \sqrt{1}$.
$\sqrt{100} = 10$ and $\sqrt{1} = 1$.
The sum is $10 – 1 = 9$.