Binomial Expansion Simplification

The initial expression given is \((x + y)^{2n+1}(x - y)^{2n+1}\). We can simplify this expression by combining the bases first, using the exponent rule \((a \cdot b)^m = a^m \cdot b^m\). \((x + y)^{2n+1}(x - y)^{2n+1} = [(x + y)(x - y)]^{2n+1}\) Now, applying the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\), to the terms inside the brackets: \([(x + y)(x - y)]^{2n+1} = (x^2 - y^2)^{2n+1}\) So, the problem is essentially about the binomial expansion of \((x^2 - y^2)^{2n+1}\).

Identifying Middle Terms

The binomial expansion of \((a+b)^N\) contains \(N+1\) terms in total. In our simplified expression, \((x^2 - y^2)^{2n+1}\), we have \(a = x^2\), \(b = -y^2\), and the power is \(N = 2n+1\). Since \(N = 2n+1\) is an odd number, the total number of terms in the expansion will be \(N+1 = (2n+1)+1 = 2n+2\). This is an even number. When an expansion has an even number of terms, say \(2k\), the middle terms are the \(k\)-th term and the \((k+1)\)-th term. In our case, the total number of terms is \(2n+2\). So, \(2k = 2n+2\), which implies \(k = n+1\). Therefore, the middle terms are the \((n+1)\)-th term and the \((n+2)\)-th term of the expansion.

Calculating Middle Terms

The formula for the general term (the \((r+1)\)-th term) in the binomial expansion of \((a+b)^N\) is: \(T_{r+1} = \binom{N}{r} a^{N-r} b^r\) For the expansion of \((x^2 - y^2)^{2n+1}\), we substitute \(a = x^2\), \(b = -y^2\), and \(N = 2n+1\) into the general term formula: \(T_{r+1} = \binom{2n+1}{r} (x^2)^{2n+1-r} (-y^2)^r\) To find the \((n+1)\)-th term, we set \(r = n\) (because \(r+1 = n+1\)): \(T_{n+1} = \binom{2n+1}{n} (x^2)^{2n+1-n} (-y^2)^n\) \(T_{n+1} = \binom{2n+1}{n} (x^2)^{n+1} (-1)^n (y^2)^n\) Simplifying the powers: \(T_{n+1} = \binom{2n+1}{n} x^{2n+2} y^{2n} (-1)^n\) To find the \((n+2)\)-th term, we set \(r = n+1\) (because \(r+1 = n+2\)): \(T_{n+2} = \binom{2n+1}{n+1} (x^2)^{2n+1-(n+1)} (-y^2)^{n+1}\) \(T_{n+2} = \binom{2n+1}{n+1} (x^2)^{n} (-1)^{n+1} (y^2)^{n+1}\) Simplifying the powers: \(T_{n+2} = \binom{2n+1}{n+1} x^{2n} y^{2n+2} (-1)^{n+1}\)

Applying the Sum Condition

The problem states that the sum of these two middle terms is zero: \(T_{n+1} + T_{n+2} = 0\) Substituting the expressions we derived: \(\binom{2n+1}{n} x^{2n+2} y^{2n} (-1)^n + \binom{2n+1}{n+1} x^{2n} y^{2n+2} (-1)^{n+1} = 0\) We utilize the property of binomial coefficients: \(\binom{N}{r} = \binom{N}{N-r}\). Applying this, we find: \(\binom{2n+1}{n} = \binom{2n+1}{(2n+1)-n} = \binom{2n+1}{n+1}\) Let \(C\) represent the value of these binomial coefficients, so \(C = \binom{2n+1}{n} = \binom{2n+1}{n+1}\). The equation now is: \(C x^{2n+2} y^{2n} (-1)^n + C x^{2n} y^{2n+2} (-1)^{n+1} = 0\) Assuming \(C \neq 0\), \(x \neq 0\), and \(y \neq 0\), we can divide the entire equation by \(C x^{2n} y^{2n}\): \(x^2 (-1)^n + y^2 (-1)^{n+1} = 0\) We can rewrite \((-1)^{n+1}\) as \((-1)^n \cdot (-1)^1 = -(-1)^n\). Substituting this gives: \(x^2 (-1)^n + y^2 [-(-1)^n] = 0\) \(x^2 (-1)^n - y^2 (-1)^n = 0\) Factor out the common term \((-1)^n\): \((-1)^n (x^2 - y^2) = 0\) Since \((-1)^n\) is always non-zero (it's either 1 or -1), for the product to be zero, the other factor must be zero: \(x^2 - y^2 = 0\) This simplifies to: \(x^2 = y^2\)

Determining the Value of (x^2 / y^2)

The question asks for the value of the ratio \(\left(\frac{x^2}{y^2}\right)\). We have established from the sum of the middle terms condition that \(x^2 = y^2\). Provided that \(y \neq 0\) (which is usually assumed in such algebraic problems unless stated otherwise), we can divide both sides of the equation \(x^2 = y^2\) by \(y^2\): \(\frac{x^2}{y^2} = \frac{y^2}{y^2}\) \(\frac{x^2}{y^2} = 1\) Therefore, the value of \(\left(\frac{x^2}{y^2}\right)\) is 1.