Solution

1. Key Concept: General Term of Binomial Expansion

The binomial theorem states that for any real numbers $a$ and $b$, and any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} {}^n C_r a^{n-r} b^r$ The $(r+1)$-th term, often denoted as $T_{r+1}$, gives the general form of each term in the expansion. It is expressed as: $T_{r+1} = {}^n C_r a^{n-r} b^r$ where $r$ is an integer ranging from $0$ to $n$.

For the given expression $\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)^{n}$, we identify $a = \sqrt{x} = x^{1/2}$ and $b = -\frac{6}{x^{3/2}} = -6x^{-3/2}$.

2. Deriving the General Term with $x$ Power

Now, we substitute these into the general term formula to find the form of $T_{r+1}$ for our specific expansion: $T_{r+1} = {}^n C_r (x^{1/2})^{n-r} (-6x^{-3/2})^r$ We simplify this expression by applying the exponent rules $(x^m)^k = x^{mk}$ and $x^m x^k = x^{m+k}$: $T_{r+1} = {}^n C_r (x^{\frac{n-r}{2}}) (-6)^r (x^{-\frac{3r}{2}})$ Group the numerical coefficient and the terms involving $x$: $T_{r+1} = {}^n C_r (-6)^r x^{\frac{n-r}{2} - \frac{3r}{2}}$ Combine the exponents of $x$: $T_{r+1} = {}^n C_r (-6)^r x^{\frac{n-r-3r}{2}}$ $T_{r+1} = {}^n C_r (-6)^r x^{\frac{n-4r}{2}}$ This is the general term of the expansion, showing its coefficient and the power of $x$.

3. Determining the Constant Term ($\alpha$) and the Value of $n$

• Finding the Constant Term ($\alpha$): A term is constant when the power of $x$ in that term is $0$. So, we set the exponent of $x$ from our general term equal to $0$: $\frac{n-4r}{2} = 0$ $n-4r = 0$ $n = 4r$ Since $r$ must be an integer (as it is the index of the term), $n$ must be a multiple of $4$. The constant term $\alpha$ is the coefficient of this term, obtained by substituting $r = \frac{n}{4}$ into the coefficient part of $T_{r+1}$: $\alpha = {}^n C_{n/4} (-6)^{n/4}$

• Using the Sum of Coefficients: The sum of all coefficients in the binomial expansion of $P(x)$ is obtained by substituting $x=1$ into the original expression. $\text{Sum of all coefficients} = \left(\sqrt{1}-\frac{6}{1^{\frac{3}{2}}}\right)^{n} = (1-6)^n = (-5)^n$ The problem states that "the sum of the coefficients of the remaining terms in the expansion is 649". "Remaining terms" refers to all terms except the constant term. Therefore: $(\text{Sum of all coefficients}) - (\text{Coefficient of constant term}) = 649$ $(-5)^n - \alpha = 649$ Substitute the expression for $\alpha$: $(-5)^n - {}^n C_{n/4} (-6)^{n/4} = 649$ We are given that $n \leq 15$. Since $n$ must be a multiple of 4, the possible values for $n$ are $4, 8, 12$. Let's test these values:

  • If $n=4$: Then $r = n/4 = 4/4 = 1$. $\alpha = {}^4 C_1 (-6)^1 = 4 \times (-6) = -24$. Check the equation: $(-5)^4 - (-24) = 625 + 24 = 649$. This matches the given condition.
  • If $n=8$: Then $r = n/4 = 8/4 = 2$. $\alpha = {}^8 C_2 (-6)^2 = \frac{8 \times 7}{2} \times 36 = 28 \times 36 = 1008$. Check the equation: $(-5)^8 - 1008 = 390625 - 1008 \neq 649$. This is incorrect.
  • If $n=12$: Then $r = n/4 = 12/4 = 3$. $\alpha = {}^{12} C_3 (-6)^3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \times (-216) = 220 \times (-216)$, which is a large negative number. This will not satisfy the equation.

  Therefore, the only valid value for $n$ is $4$, and the constant term $\alpha = -24$.

  Tip: Always check all given constraints ($n \leq 15$ in this case) and test values systematically to ensure correctness.

4. Finding the Coefficient of $x^{-n}$

We need to find the coefficient of $x^{-n}$. Since we found $n=4$, this means we need the coefficient of $x^{-4}$. From the general term $T_{r+1} = {}^n C_r (-6)^r x^{\frac{n-4r}{2}}$, we set the exponent of $x$ to $-4$: $\frac{n-4r}{2} = -4$ Substitute $n=4$ into this equation: $\frac{4-4r}{2} = -4$ Multiply both sides by $2$: $4-4r = -8$ Subtract $4$ from both sides: $-4r = -12$ Divide by $-4$: $r = 3$ Now, we substitute $n=4$ and $r=3$ into the coefficient part of the general term to find the coefficient of $x^{-4}$: $\text{Coefficient of } x^{-4} = {}^4 C_3 (-6)^3$ Calculate the binomial coefficient and the power: ${}^4 C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4$ $(-6)^3 = -216$ So, the coefficient of $x^{-4} = 4 \times (-216) = -864$.

5. Calculating $\lambda$

The problem states that the coefficient of $x^{-n}$ is equal to $\lambda \alpha$. We have calculated:

• Coefficient of $x^{-n}$ (which is $x^{-4}$) $= -864$.
• $\alpha = -24$. Substitute these values into the given relationship: $-864 = \lambda \times (-24)$ To find $\lambda$, divide both sides by $-24$: $\lambda = \frac{-864}{-24}$ $\lambda = 36$

Common Mistake: Be careful with signs, especially when dealing with negative bases raised to powers. Also, ensure accurate calculation of binomial coefficients.

6. Summary and Key Takeaway

By systematically using the general term of the binomial expansion, applying the conditions for the constant term and the sum of remaining coefficients, we determined that $n=4$ and the constant term $\alpha=-24$. Subsequently, we found the term corresponding to $x^{-4}$ and its coefficient to be $-864$. Finally, equating this to $\lambda \alpha$ yielded $\lambda=36$. This problem emphasizes the importance of a clear understanding of the general term, properties of binomial coefficients, and careful algebraic manipulation.

The final answer is $\boxed{36}$.