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

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:

$119 - 1 = 118.$