Q8) Find the term contains \( x^{4} \) in the expansion of \( \left(x^{2}-\frac{1}{x^{2}}\right)^{12} \) according to the descending powers of \( x \), then find the ratio between the coefficient of this term and the middle term.

To find the term that contains \( x^{4} \) in the expansion of \( \left(x^{2} - \frac{1}{x^{2}}\right)^{12} \), we can apply the binomial theorem. The general term in the expansion can be written as: \[ T_k = \binom{12}{k} (x^{2})^{12-k} \left(-\frac{1}{x^{2}}\right)^k = \binom{12}{k} (-1)^k x^{2(12-k)} x^{-2k} \] This simplifies to: \[ T_k = \binom{12}{k} (-1)^k x^{24 - 4k} \] We need \( 24 - 4k = 4 \) to find the term that contains \( x^4 \): \[ 24 - 4k = 4 \implies 20 = 4k \implies k = 5 \] Now, substituting \( k = 5 \) into the expression for \( T_k \): \[ T_5 = \binom{12}{5} (-1)^5 x^{4} = -\binom{12}{5} x^{4} \] Now, we need the coefficient of this term: \[ \text{Coefficient} = -\binom{12}{5} \] Next, let's find the middle term of the expansion of \( \left(x^{2} - \frac{1}{x^{2}}\right)^{12} \). Since there are 13 terms in total, the middle term (the 7th term) will be: \[ T_6 = \binom{12}{6} (x^{2})^{12-6} \left(-\frac{1}{x^{2}}\right)^6 = \binom{12}{6} (-1)^6 x^{12 - 12} = \binom{12}{6} \] The coefficient of the middle term is: \[ \text{Coefficient of middle term} = \binom{12}{6} \] Now, we want to find the ratio of the coefficients of the term containing \( x^{4} \) and the middle term: \[ \text{Ratio} = \frac{-\binom{12}{5}}{\binom{12}{6}} = -\frac{\binom{12}{5}}{\binom{12}{6}} \] Using the relationship \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \): \[ \frac{\binom{12}{5}}{\binom{12}{6}} = \frac{6}{12 - 5} = \frac{6}{7} \] Thus, the ratio is: \[ \text{Ratio} = -\frac{6}{7} \] Final Answer: The term containing \( x^{4} \) has a coefficient of \( -\binom{12}{5} \) and the middle term has a coefficient of \( \binom{12}{6} \), giving a ratio of \( -\frac{6}{7} \).