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 (\frac{x^{2}}{y^{2}})?

Explanation

The expression simplifies to ((x+y)(x-y))^{2n+1} = (x^2 - y^2)^{2n+1}. The middle terms are the (n+1)-th and (n+2)-th terms. Their sum is \binom{2n+1}{n}(x^2)^{n+1}(-y^2)^n + \binom{2n+1}{n+1}(x^2)^n(-y^2)^{n+1} = 0. Since \binom{2n+1}{n} = \binom{2n+1}{n+1}, we divide the entire equation by \binom{2n+1}{n}(x^2)^n(-y^2)^n to get x^2 - y^2 = 0, leading to \frac{x^2}{y^2} = 1.

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