Question

In the expansion of \[ (1 + x)(1 - x^2) \left( 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} \right)^5, \quad x \neq 0, \]the sum of the coefficients of \( x^3 \) and \( x^{-13} \) is equal to ____

Answer 1

Correct Answer: 118

Rewriting the given expression:

\[ (1 + x)(1 - x^2) \left( 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} \right)^5, \quad x \neq 0, \]

Expanding:

\[ (1 + x)^2(1 - x)^{17} \]

To find the coefficient of \( x^2 \) in the expansion:

Coeff of \( x^2 \) = combination and calculation shown = \( 17 \)

Similarly, for \( x^{-13} \):

\( (1 + x)(1 - x^2) \left( 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} \right)^5\) 

\(= (1 + x)(1 - x^2) \left( \frac{(1 + x)^3}{x^3} \right)^5 \)

\(= (1 + x)^2(1 - x^2) \frac{(1 + x)^{15}}{x^{15}}\)

\(= \frac{(1 + x)^{17} - x(1 + x)^{17}}{x^{15}}\)

\(= \text{coeff}\left( x^3 \right) \text{ in the expansion of } (1 + x)^{17} - x(1 + x)^{17} = 0 - 1 = -1\)

Coeff \( x^{-13} \) = Coeff \( x^2 \) in \( (1 + x)^{17} - x(1 + x)^{17} \)

\(= \binom{17}{2} - \binom{17}{1}= 17 \times 8 - 17  = 119\)

Hence Answer:

\[ 119 - 1 = 118. \]