In the expansion of (1 + x)(1 - x ^ 2 ) (1 + (3)/(x) + 3 x ^ 2 + 1...

Given,

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

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

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

$= \dfrac{{\left(1 + x\right)}^{17}\left(1 - x\right)}{{x}^{15}}$

$= \dfrac{{\left(1 + x\right)}^{17}}{{x}^{15}} - \dfrac{{\left(1 + x\right)}^{17}}{{x}^{14}}$

Now for coefficient of ${x}^{3}$, we will find coefficient of ${x}^{18} in expansion of${\left(1 + x\right)}^{17}$which is not possible and${x}^{17} in expansion of $- {\left(1 + x\right)}$^{17}$which will be$- C17 17 = - 1$</p><p>And for coefficient of${x}^{- 13}$we will coefficient of${x}^{2} in expansion of ${\left(1 + x\right)}$^{17}$which will be$C2 17$and coefficient of$x$in expansion of$- {\left(1 + x\right)}^{17} which will be $- C1 17$, so the coefficient of ${x}^{- 13}$ will be $C2 17 - C1 17 = 136 - 17 = 119$

Hence, the sum will be $119 - 1 = 118$