Question

In the expansion of \(\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}, x>0\), if the term independent of \(x\) is \((221)k\), then \(k\) is equal to:

• A. 84
• B. 78
• C. 168
• D. 198

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the term independent of \(x\) in a binomial expansion, write out the general term, collect all powers of \(x\), and set the final exponent equal to zero to solve for the term index \(r\).
Step 2: Key Formula or Approach:
General term in \((a + b)^n\):
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \] Step 3: Detailed Explanation:
The given binomial is \((9x - \frac{1}{3}x^{-1/2})^{18}\).
\[ T_{r+1} = \binom{18}{r} (9x)^{18-r} \left( -\frac{1}{3}x^{-1/2} \right)^r \] Combine the exponents of \(x\):
\[ x^{18 - r} \cdot x^{-r/2} = x^{18 - 3r/2} \] For the term to be independent of \(x\), set the exponent to 0:
\[ 18 - \frac{3r}{2} = 0 \implies r = 12 \] Calculate the coefficient for \(r = 12\):
\[ T_{13} = \binom{18}{12} (9)^{6} \left(-\frac{1}{3}\right)^{12} \] Since \(9^6 = (3^2)^6 = 3^{12}\) and \((-1/3)^{12} = 3^{-12}\):
\[ T_{13} = \binom{18}{12} \cdot 3^{12} \cdot 3^{-12} = \binom{18}{12} \] Compute \(\binom{18}{12} = \binom{18}{6}\):
\[ \binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564 \] Equating to the given expression:
\[ 221k = 18564 \implies k = \frac{18564}{221} = 84 \] Step 4: Final Answer:
The value of \(k\) is 84.