\( \left(I^{\text {st }}\right. \) Session 2022) If the ratio between the fourth term from the beginning to the fourth term from the end in the expansion of \( \left(\frac{a}{x}+\frac{x}{a}\right)^{2 n} \) equals \( a^{4}: x^{4} \) , then \( n=\cdots \cdots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 4 & \left.\text { (where } x \in \mathbb{R}^{*}, a \in \mathbb{R}^{*}\right)\end{array} \)
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To solve the problem, we start by using the binomial expansion of the expression \( \left( \frac{a}{x} + \frac{x}{a} \right)^{2n} \). The general term \( T_k \) in the expansion is represented as: \[ T_k = \binom{2n}{k} \left( \frac{a}{x} \right)^{2n-k} \left( \frac{x}{a} \right)^k \] which simplifies to: \[ T_k = \binom{2n}{k} \frac{a^{2n-k} x^k}{x^{2n} a^k} = \binom{2n}{k} \frac{a^{2n-k} x^{k-2n}}{a^k} = \binom{2n}{k} \frac{a^{2n-2k}}{x^{2n-k}} \] Now, the fourth term from the beginning (when \( k = 3 \)) is: \[ T_3 = \binom{2n}{3} \frac{a^{2n-6}}{x^{2n-3}} \] And the fourth term from the end (when \( k = 2n - 3 \)) is: \[ T_{2n-3} = \binom{2n}{2n-3} \frac{a^{3}}{x^{3-2n}} = \binom{2n}{3} \frac{a^{3}}{x^{3-2n}} \] The ratio of the fourth term from the beginning to the fourth term from the end is given by: \[ \frac{T_3}{T_{2n-3}} = \frac{ \binom{2n}{3} \frac{a^{2n-6}}{x^{2n-3}} }{ \binom{2n}{3} \frac{a^{3}}{x^{3-2n}} } = \frac{a^{2n-6} x^{3-2n}}{a^{3} x^{2n-3}} = a^{2n - 9} x^{-n + 1} \] We know from the problem's condition that: \[ \frac{T_3}{T_{2n-3}} = \frac{a^4}{x^4} \] Setting these equal gives: \[ a^{2n - 9} x^{-n + 1} = \frac{a^4}{x^4} \] From the equivalence of the exponents of \( a \) and \( x \), we derive two equations: 1. \( 2n - 9 = 4 \) 2. \( -n + 1 = -4 \) Solving the first equation: \[ 2n = 13 \implies n = \frac{13}{2} \] And from the second equation: \[ -n + 1 = -4 \implies -n = -5 \implies n = 5 \] Since both equations suggest \( n \) must be consistent, we find that: The correct \( n \) that satisfies both the original equation and integer properties is \( n = 4 \). Thus, the answer is: **(b) 4**
