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 coefficients of x^3 and x^-13 is equal to __________.

(1+x)(1-x^2)(1+(3 / x)+(3 / x^2)+(1 / x^3))^5 \ = (1+x)(1-x^2)((1+(1 / x))^3)^5 \ = frac(1+x)^2(1-x)(1+x)^15x^15 \ = frac(1+x)^17-x(1+x)^17x^15 \ To find the coeff(x^3) in the expansion, \ we find the coeff(x^18) in: (1+x)^17-x(1+x)^17 \ = 0 - 1 \ = -1 \ To find the coeff(x^-13) in the expansion, \ we find the coeff(x^2) in: (1+x)^17-x(1+x)^17 \ = binom172 - binom171 \ = (17 x 8) - 17 \ = 17 x 7 \ = 119 \ Hence, Answer = 119 - 1 = 118

In the expansion of 1x1x21frac 3xfrac 3x2frac 1x35 x ne... | (Mathematics)

\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}

\text{To find the } \text{coeff}(x^3) \text{ in the expansion,}

\\ \text{ we find the coeff}(x^{18}) \text{ in:} \\

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

\text{To find the } \text{coeff}(x^{-13}) \text{ in the expansion,}

\\ \text{ we find the coeff}(x^2) \text{ in:} \\

\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ &= \binom{17}{2} - \binom{17}{1} \\ &= (17 \times 8) - 17 \\ &= 17 \times 7 \\ &= 119 \end{aligned}

\text{Hence, Answer } = 119 - 1 = 118

Explanation

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