Question

The coefficient of \( x^3 \) in the power series expansion of \( \frac{1+4x-3x^2}{(1+3x)^3} \) is

• A. \( -27 \)
• B. \( 27 \)
• C. \( 153 \)
• D. \( -153 \) Correct Answer

Hint:

To find the coefficient of \(x^k\) in an expression like \( P(x) (1+ax)^{-n} \), where \(P(x)\) is a polynomial, expand \( (1+ax)^{-n} \) up to the \(x^k\) term using the binomial theorem for negative or fractional indices: \( (1+Y)^{-n} = \sum_{j=0}^{\infty} \binom{-n}{j} Y^j = \sum_{j=0}^{\infty} (-1)^j \binom{n+j-1}{j} Y^j \). Then multiply by \(P(x)\) and collect the terms that result in \(x^k\).

Answer 1

The Correct Option is A

Step 1: Rewrite the expression.
The expression is \( (1+4x-3x^2)(1+3x)^{-3} \).

Step 2: Expand \( (1+3x)^{-3} \) using the binomial theorem for a negative integer index.
The formula is \( (1+Y)^{-n} = 1 - nY + \frac{n(n+1)}{2!}Y^2 - \frac{n(n+1)(n+2)}{3!}Y^3 + \dots \).
Here, \( Y = 3x \) and \( n=3 \).
We need terms up to \(x^3\).
\[ (1+3x)^{-3} = 1 - 3(3x) + \frac{3(3+1)}{2}(3x)^2 - \frac{3(3+1)(3+2)}{6}(3x)^3 + \dots \] \[ = 1 - 9x + \frac{3 \cdot 4}{2}(9x^2) - \frac{3 \cdot 4 \cdot 5}{6}(27x^3) + \dots \] \[ = 1 - 9x + 6(9x^2) - 10(27x^3) + \dots \] \[ = 1 - 9x + 54x^2 - 270x^3 + \dots \]
Step 3: Multiply \( (1+4x-3x^2) \) by the expansion of \( (1+3x)^{-3} \).
We are looking for the coefficient of \(x^3\).
The terms that produce \(x^3\) are: \begin{itemize} \item \( 1 \times (-270x^3) = -270x^3 \) \item \( (4x) \times (54x^2) = 216x^3 \) \item \( (-3x^2) \times (-9x) = 27x^3 \) \end{itemize}
Step 4: Sum the coefficients of these \(x^3\) terms.
Coefficient of \(x^3\) = \( -270 + 216 + 27 \).
\[ -270 + 216 = -54 \] \[ -54 + 27 = -27 \] The coefficient of \( x^3 \) is -27.
This matches option (1).