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Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1+x)(2-x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get $\binom{17}{2}-\binom{17}{1}$ but in this book it's just $\binom{17}{2}$.

Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1-x)(1+x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get $\binom{17}{2}-\binom{17}{1}$ but in this book it's just $\binom{17}{2}$.

Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1+x)(2-x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get $\binom{17}{2}-\binom{17}{1}$ but in book it's just $\binom{17}{2}$.

Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1+x)(2-x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get $\binom{17}{2}-\binom{17}{1}$ but in this book it's just $\binom{17}{2}$.

Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1+x)(2-x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get ${}^{17}C_2 - {}^{17}C_1$ but in book it's just $^{17}C_2$.

Consider the following question

In the expansion of $(1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5$, $x \ne 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to ____.

what I tried: To find the coefficient of $x^{-13}$ we need to get $x^2$ from $(1+x)(2-x)^{17}$ so we have to make cases. We have two cases: first we take $x^2$ from the equation $(1+x)^{17}$ and $x^0$ from $(1-x)$ and in the second case will take $x^1$ from both the brackets and thus if we add the coefficients of $x^{-13}$ from both cases to get $\binom{17}{2}-\binom{17}{1}$ but in book it's just $\binom{17}{2}$.