In a binomial expansion of (x + y)^(2n+1) (x – y)^(2n+1), the sum of middle terms is zero. What is the value of (x²/y²)?

(x+y)^(2n+1)(x–y)^(2n+1) = (x² – y²)^(2n+1). This has 2n+2 terms, so two middle terms: T_(n+1) and T_(n+2). T_(n+1) = C(2n+1, n)(x²)^(n+1)(–y²)ⁿ and T_(n+2) = C(2n+1, n+1)(x²)ⁿ(–y²)^(n+1). Sum = C(2n+1,n)x^(2n+2)(–y²)ⁿ + C(2n+1,n+1)x^(2n)(–y²)^(n+1). Since C(2n+1,n) = C(2n+1,n+1), sum = C(2n+1,n)(–y²)ⁿ[x^(2n+2) – x^(2n)y²] = 0. This gives x^(2n+2) = x^(2n)y², so x² = y², hence x²/y² = 1.