Question

In the expansion of \[ \left(9x-\frac{1}{3\sqrt{x}}\right)^{18}, \] if the coefficient of the term independent of \(x\) is \(221k\), then the value of \(k\) is:

• A. \(82\)
• B. \(83\)
• C. \(84\)
• D. \(86\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks to find a constant \(k\) by identifying the term independent of \(x\) in a given binomial expansion and equating its coefficient to \(221k\).
Step 2: Key Formula or Approach:
The general term in the expansion of \((a + b)^{n}\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^{r} \]
For the term to be independent of \(x\), the total power of \(x\) must be zero.
Step 3: Detailed Explanation:
In the given expansion \((9x - \frac{1}{3\sqrt{x}})^{18}\), we have \(a = 9x\), \(b = -\frac{1}{3}x^{-1/2}\), and \(n = 18\).
The general term is:
\[ T_{r+1} = \binom{18}{r} (9x)^{18-r} \left( -\frac{1}{3\sqrt{x}} \right)^{r} \]
\[ T_{r+1} = \binom{18}{r} (3^{2})^{18-r} x^{18-r} (-1)^{r} (3^{-1})^{r} x^{-r/2} \]
\[ T_{r+1} = \binom{18}{r} (-1)^{r} 3^{36-2r} 3^{-r} x^{18-r-r/2} \]
\[ T_{r+1} = \binom{18}{r} (-1)^{r} 3^{36-3r} x^{18 - \frac{3r}{2}} \]
For the term independent of \(x\), the power of \(x\) is \(0\):
\[ 18 - \frac{3r}{2} = 0 \Rightarrow 18 = \frac{3r}{2} \Rightarrow r = 12 \]
The coefficient is:
\[ \text{Coeff} = \binom{18}{12} (-1)^{12} 3^{36 - 3(12)} = \binom{18}{6} \cdot 1 \cdot 3^{0} = \binom{18}{6} \]
Calculating \(\binom{18}{6}\):
\[ \binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564 \]
Given that this coefficient is \(221k\):
\[ 221k = 18564 \Rightarrow k = \frac{18564}{221} = 84 \]
Step 4: Final Answer:
The value of \(k\) is 84.