UPSC CDS I 2024 Math Paper

Concept:The given series is a telescoping sum where each fraction can be rewritten as a difference of two reciprocals, causing most terms to cancel.Explanation:Observe that the numerator of each term is the difference of squares of consecutive integers.For the $n$th term (starting with $n=1$): $\frac{(n+1)^2 - n^2}{n^2 (n+1)^2}$ simplifies to $\frac{1}{n^2} - \frac{1}{(n+1)^2}$.Applying this to the series: $\frac{3}{1^2 \times 2^2} = \frac{2^2 - 1^2}{1^2 \times 2^2} = \frac{1}{1^2} - \frac{1}{2^2}$.Similarly, $\frac{5}{2^2 \times 3^2} = \frac{1}{2^2} - \frac{1}{3^2}$, $\frac{7}{3^2 \times 4^2} = \frac{1}{3^2} - \frac{1}{4^2}$, and so on.Adding all terms, intermediate terms cancel: $1 - \frac{1}{2^2} + \frac{1}{2^2} - \frac{1}{3^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots$ up to infinity.The sum telescopes to $1 - \frac{1}{\infty^2}$, which equals $1$.Answer:$1$