1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T43f,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"In the expansion of 1x1x21frac 3xfrac 3x2frac 1x35 x neq 0 t","item":"https://www.divinejee.com/question/in-the-expansion-of-1-x1x21-frac-3x-frac-3x2-frac-1x35-x-neq-0-the-sum-of-the-coefficients-of-x3"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question In the expansion of 1x1x21frac 3xfrac 3x2frac 1x35 x neq 0 t on Topic","text":"

In the expansion of $(1+x)\\left(1-x^2\\right)\\left(1+\\frac{3}{x}+\\frac{3}{x^2}+\\frac{1}{x^3}\\right)^5, x \\neq 0$, the sum of the coefficients of $x^...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/in-the-expansion-of-1-x1x21-frac-3x-frac-3x2-frac-1x35-x-neq-0-the-sum-of-the-coefficients-of-x3"}}}]}20:T4e7,

$$\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \end{aligned}$$

$$=\operatorname{coeff}\left(\mathrm{x}^3\right)$$ in the expansion $$\approx \operatorname{coeff}\left(\mathrm{x}^{18}\right)$$ in

$$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =0-1 \\ & =-1 \end{aligned}$$

$$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$$ in the expansion $$\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$$ in

$$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =\left(\begin{array}{c} 17 \\ 2 \end{array}\right)-\left(\begin{array}{c} 17 \\ 1 \end{array}\right) \\ & =17 \times 8-17 \\ & =17 \times 7 \\ & =119 \end{aligned}$$

Hence Answer $$=119-1=118$$