JEE Main 202431 Jan, Morning shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002In the expansion of (1+x)(1−x2)(1+3x+3x2+1x3)5,x≠0(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0(1+x)(1−x2)(1+x3+x23+x31)5,x=0, the sum of the coefficients of x3x^3x3 and x−13x^{-13}x−13 is equal to __________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution