UPSC NDA I 2024 Mathematics Paper

Concept:The product $(x+y)^{2n+1}(x-y)^{2n+1}$ simplifies to $(x^2 - y^2)^{2n+1}$.For a binomial expansion with even power, there are two middle terms whose sum is given to be zero.Explanation:Combine the two binomials: $(x+y)^{2n+1}(x-y)^{2n+1} = \left[(x+y)(x-y)\right]^{2n+1} = (x^2 - y^2)^{2n+1}$.The expansion of $(x^2 - y^2)^{2n+1}$ has $2n+2$ terms (even).Thus, there are two middle terms: the $(n+1)$th and $(n+2)$th terms.Using the binomial theorem, the general term is ${}^{2n+1}C_r \, (x^2)^{2n+1-r} \, (-y^2)^r$.For $r = n$ (first middle term) and $r = n+1$ (second middle term), the terms are:$T_{n+1} = {}^{2n+1}C_n \, (x^2)^{n+1} \, (-y^2)^n$ and $T_{n+2} = {}^{2n+1}C_{n+1} \, (x^2)^{n} \, (-y^2)^{n+1}$.Since ${}^{2n+1}C_n = {}^{2n+1}C_{n+1}$, the sum of the two middle terms is:${}^{2n+1}C_n \left[ (x^2)^{n+1} (-y^2)^n + (x^2)^n (-y^2)^{n+1} \right] = {}^{2n+1}C_n \, (x^2)^n (y^2)^n \left( -x^2 + y^2 \right)$.Setting this sum to zero gives $-x^2 + y^2 = 0$, i.e., $x^2 = y^2$.Therefore, $\frac{x^2}{y^2} = 1$.Answer:$1$ (Option A).