Question

If the expansion of \[ \left(\frac{1+15x}{1-3x}\right) \] is valid and the coefficient of \(x^3\) in its expansion is \(k(3^3)\), then \(k=\)

• A. \(7\)
• B. \(6\)
• C. \(12\)
• D. \(13\)

Hint:

For coefficient problems, identify every possible multiplication that can produce the required power of \(x\). Missing even one contribution leads to an incorrect answer.

Answer 1

The Correct Option is B

Step 1: Expand the denominator as a series.
For $|3x| < 1$, $\dfrac{1}{1-3x} = 1 + 3x + 3^2x^2 + 3^3x^3 + \cdots$, a simple geometric series.
Step 2: Write the full product.
\[ \frac{1+15x}{1-3x} = (1 + 15x)\big(1 + 3x + 9x^2 + 27x^3 + \cdots\big). \]
Step 3: Hunt for $x^3$ terms.
An $x^3$ can come from $1 \times 27x^3$ or from $15x \times 9x^2$.
Step 4: Compute each contribution.
First piece: $1 \times 27 = 27$. Second piece: $15 \times 9 = 135$.
Step 5: Add them.
Coefficient of $x^3 = 27 + 135 = 162$.
Step 6: Solve for $k$.
Given this equals $k\cdot 3^3 = 27k$, we get $k = \dfrac{162}{27} = 6$, which is option (B).
\[ \boxed{k = 6} \]