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 \(k=\)
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 \(k=\)
Show 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.
Updated On: Jun 9, 2026
- \(7\)
- \(6\)
- \(12\)
- \(13\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For \(|3x|<1\),
\[
\frac1{1-3x}
=
1+3x+3^2x^2+3^3x^3+\cdots.
\]
This is the standard geometric progression expansion
\[
\frac1{1-r}=1+r+r^2+r^3+\cdots.
\]
Step 1: Expand the denominator. \[ \frac{1+15x}{1-3x} = (1+15x) \left( 1+3x+9x^2+27x^3+\cdots \right). \]
Step 2: Find all contributions to the coefficient of \(x^3\). The \(x^3\)-term can arise in two ways: \[ 1\times 27x^3 \] giving contribution \[ 27. \] Also, \[ 15x\times 9x^2 \] giving contribution \[ 135. \]
Step 3: Add the contributions. Hence coefficient of \(x^3\) is \[ 27+135=162. \]
Step 4: Compare with \(k(3^3)\). Since \[ 162=k(27), \] \[ k=\frac{162}{27}=6. \] Therefore, \[ \boxed{6}. \] Hence the correct option is \(\boxed{(B)}\).
Step 1: Expand the denominator. \[ \frac{1+15x}{1-3x} = (1+15x) \left( 1+3x+9x^2+27x^3+\cdots \right). \]
Step 2: Find all contributions to the coefficient of \(x^3\). The \(x^3\)-term can arise in two ways: \[ 1\times 27x^3 \] giving contribution \[ 27. \] Also, \[ 15x\times 9x^2 \] giving contribution \[ 135. \]
Step 3: Add the contributions. Hence coefficient of \(x^3\) is \[ 27+135=162. \]
Step 4: Compare with \(k(3^3)\). Since \[ 162=k(27), \] \[ k=\frac{162}{27}=6. \] Therefore, \[ \boxed{6}. \] Hence the correct option is \(\boxed{(B)}\).
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