In the expansion of (1+x)(1-x^2)(1+(3)/(x)+3x^2+1x^3)^5,x 0, the sum of the coefficients o…

Question

In the expansion of $(1+x)(1-{x}^{2}){(1+\frac{3}{x}+\frac{3}{{x}^{2}}+\frac{1}{{x}^{3}})}^{5},x
e 0$, the sum of the coefficients of ${x}^{3}$ and ${x}^{-13}$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 118

Accepted Answer

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

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

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

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

$\begin{gathered} (1+x{)}^{17}-x(1+x{)}^{17} \ =(\begin{gathered} 17 \ 2)-(17 \ 1 \end{gathered}) \ =17	imes 8-17 \ =17	imes 7 \ =119 \end{gathered}$

Hence Answer $=119-1=118$

Solutions · 1