Question

If \(\left(\frac{3^6}{4^4}\right)k\) is the term independent of \(x\) in the binomial expansion of \(\left(\frac{x}{4} - \frac{12}{x^2}\right)^{12}\), then \(k\) is equal to ___________
 

Hint:

In \((ax^p + bx^q)^n\), the term independent of \(x\) occurs at \(r = \frac{np}{p-q}\). Here, \(r = \frac{12(1)}{1 - (-2)} = \frac{12}{3} = 4\).

Answer 1

Correct Answer: 55

Step 1: Understanding the Concept:
In the binomial expansion of \((a+b)^n\), the general term is given by \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).
To find the term independent of \(x\), we find the value of \(r\) for which the net power of \(x\) in the general term is zero.
Step 2: Key Formula or Approach:
For the expansion of \(\left(\frac{x}{4} - \frac{12}{x^2}\right)^{12}\), the general term is:
\[ T_{r+1} = \binom{12}{r} \left(\frac{x}{4}\right)^{12-r} \left(-\frac{12}{x^2}\right)^r \]
Step 3: Detailed Explanation:
Simplify the powers of \(x\):
\[ T_{r+1} = \binom{12}{r} \left(\frac{1}{4}\right)^{12-r} x^{12-r} (-12)^r x^{-2r} \]
\[ T_{r+1} = \binom{12}{r} \left(\frac{1}{4}\right)^{12-r} (-12)^r x^{12-3r} \]
For the term independent of \(x\), set the exponent of \(x\) to zero:
\[ 12 - 3r = 0 \implies r = 4 \]
The independent term is \(T_5\):
\[ T_5 = \binom{12}{4} \left(\frac{1}{4}\right)^{12-4} (-12)^4 = \binom{12}{4} \frac{1}{4^8} \cdot 3^4 \cdot 4^4 = \binom{12}{4} \frac{3^4}{4^4} \]
We are given that this term is equal to \(\left(\frac{3^6}{4^4}\right)k\):
\[ \binom{12}{4} \frac{3^4}{4^4} = \frac{3^6}{4^4} k \]
\[ \binom{12}{4} = 3^2 k \implies 9k = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} \]
\[ 9k = 495 \implies k = 55 \]
Step 4: Final Answer:
The value of \(k\) is 55.