UPSC CDS 1 2022 Math Paper

Concept:Rationalize each denominator to convert the sum into a telescoping series.Explanation:Take the first term: $\frac{1}{5\sqrt{4} + 4\sqrt{5}}$.Multiply numerator and denominator by $5\sqrt{4} - 4\sqrt{5}$.This gives $\frac{5\sqrt{4} - 4\sqrt{5}}{(5\sqrt{4})^2 - (4\sqrt{5})^2} = \frac{5\sqrt{4} - 4\sqrt{5}}{100 - 80}$.Simplify: $\frac{5\sqrt{4}}{20} - \frac{4\sqrt{5}}{20} = \frac{\sqrt{4}}{4} - \frac{\sqrt{5}}{5}$.Since $\frac{\sqrt{4}}{4} = \frac{1}{\sqrt{4}}$ and $\frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$, the term becomes $\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}$.Similarly, each term $\frac{1}{(n+1)\sqrt{n} + n\sqrt{n+1}}$ simplifies to $\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$.Applying this pattern, the sum becomes:$\left(\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}\right) + \left(\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{6}}\right) + \left(\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{7}}\right) + \left(\frac{1}{\sqrt{7}} - \frac{1}{\sqrt{8}}\right) + \left(\frac{1}{\sqrt{8}} - \frac{1}{\sqrt{9}}\right)$.All intermediate terms cancel, leaving $\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{9}}$.Now $\frac{1}{\sqrt{4}} = \frac{1}{2}$ and $\frac{1}{\sqrt{9}} = \frac{1}{3}$, so the result is $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.Answer:$\frac{1}{6}$