In the expansion of (1+x)(1-x^2)(1+ 3 x+ 3 x^2+ 1 x^3)^5,x≠ 0, the sum of the coefficient of x^3 and x^-13 is equal to

Given,

$$(1+x)(1-{x}^{2}){(1+\frac{3}{x}+\frac{3}{{x}^{2}}+\frac{1}{{x}^{3}})}^{5}$$(1+x)(1−x2)(1+x3​+x23​+x31​)5

$$={(1+x)}^{2}(1-x){(\frac{{x}^{3}+3{x}^{2}+3x+1}{{x}^{3}})}^{5}$$=(1+x)2(1−x)(x3x3+3x2+3x+1​)5

$$={(1+x)}^{2}(1-x){(\frac{{(1+x)}^{3}}{{x}^{3}})}^{5}$$=(1+x)2(1−x)(x3(1+x)3​)5

$$=\frac{{(1+x)}^{17}(1-x)}{{x}^{15}}$$=x15(1+x)17(1−x)​

$$=\frac{{(1+x)}^{17}}{{x}^{15}}-\frac{{(1+x)}^{17}}{{x}^{14}}$$=x15(1+x)17​−x14(1+x)17​

Now for coefficient of ${x}^{3}$x3, we will find coefficient of ${x}^{18}$x18 in expansion of ${(1+x)}^{17}$(1+x)17 which is not possible and ${x}^{17}$x17 in expansion of $-{(1+x)}^{17}$−(1+x)17 which will be $-C1717=-1$−C1717=−1

And for coefficient of ${x}^{-13}$x−13 we will coefficient of ${x}^{2}$x2 in expansion of ${(1+x)}^{17}$(1+x)17 which will be $C217$C217 and coefficient of $x$x in expansion of $-{(1+x)}^{17}$−(1+x)17 which will be $-C117$−C117, so the coefficient of ${x}^{-13}$x−13 will be $C217-C117=136-17=119$C217−C117=136−17=119

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