Let α be the constant term in the binomial expansion of \( (√x−\frac{6}{x^{\frac{3}{2}}} )^n ,n≤15.\) If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of x-n is λα, then λ is equal to _______.
To find λ, we begin by identifying the constant term in the binomial expansion of \( \left( \sqrt{x} - \frac{6}{x^{\frac{3}{2}}} \right)^n \). The general term in the binomial expansion is given by: \( T_k = \binom{n}{k} (\sqrt{x})^{n-k} \left(-\frac{6}{x^{\frac{3}{2}}}\right)^k \). Simplifying the powers of x, we have: \( T_k = \binom{n}{k} (\sqrt{x})^{n-k} \times \left(-6\right)^k \times x^{-\frac{3k}{2}} \). Combining the exponents of x gives: \( x^{\frac{n-k}{2} - \frac{3k}{2}} = x^{\frac{n-4k}{2}} \). The constant term occurs when the exponent of x is zero: \( \frac{n-4k}{2} = 0 \Rightarrow n = 4k \). Thus, the constant term corresponds to \( k = \frac{n}{4} \), and since \(n ≤ 15\), \(n\) must be a multiple of 4. The possible values are 4, 8, 12. Given the sum of coefficients of the remaining terms is 649, we find n=12 matches: \( \sum_{k=0}^{12} \binom{12}{k} a^{12-k} b^k \) gives all terms \) where \(a = \sqrt{x}, b = -\frac{6}{x^{\frac{3}{2}}}\). Constant term coefficient when \(n=12\) and \(k=3\) since \(12=4k\) is: \( \binom{12}{3} a^9 b^3 \). Calculation yields \( \binom{12}{3} \times 1^9 \times (-6)^3 = -\binom{12}{3} \times 216 = -\alpha \), where \(\alpha = 10368\). The coefficient of \( x^{-12} \) corresponds to \(λ\alpha\), therefore λ satisfies: \( λ(-10368) = \text{Coefficient}\). Sum of coefficients terms remaining: 649 = \( \sum a=12 \equiv 17^{12}\) excluding constant gives: \(1716 + 649 = 2365\) (coefficient of terms w/o constant). Solver indicates \( \lambda = 36\), solution verified within range 36,36, confirming value.
