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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 $(1-x)(1+x)^{17}$ 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 $\binom{17}{2}$.

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  • $\begingroup$ Welcome to StackExchange. I tried to improve the formatting of your question using MathJax. However the end of your question is difficult to understand and I did not modify it. Please fix the end of your question otherwise it will likely be closed. $\endgroup$ Commented Oct 10 at 16:51
  • $\begingroup$ what do you mean by ⁷C2 -¹⁷C1 and ¹⁷C2? $\endgroup$ Commented Oct 10 at 16:51
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    $\begingroup$ @DanielSmania likely $\binom{17}{2}-\binom{17}{1}$ instead of $^{17}C_2$ and similar. $\endgroup$ Commented Oct 10 at 16:53
  • $\begingroup$ Potential hint: The expression factors into $\frac{(x + 1)^{17}(1 - x)}{x^{15}}$. $\endgroup$ Commented Oct 10 at 16:55
  • $\begingroup$ The original expression equals to $\dfrac{(1-x)(1+x)^{17}}{x^{15}}$, so $-1$ for $x^3$ and $\mathrm{C}_{17}^2-\mathrm{C}_{17}^1$ for $x^{-13}$($x^{18}$ and $x^2$ on the numerator separately). I think you made some miscalculation in the description "$2-x$" and somewhere. $\endgroup$ Commented Oct 10 at 16:55

2 Answers 2

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Let $f(x) = (1 + x)(1 - x^2)\left(1 + \dfrac{3}{x} + \dfrac{3}{x^2} + \dfrac{1}{x^3}\right)^5= \dfrac{(1 - x)(x+1)^{17}}{x^{15}}$.

By the binomial theorem,

$$f(x) = x^{-15}(1-x) \sum_{k=0}^{17} {17 \choose k}x^k1^{17-k}$$ $$= (x^{-15}-x^{-14}) \sum_{k=0}^{17} {17 \choose k}x^k$$ $$= \sum_{k=0}^{17} {17 \choose k}(x^{k-15}-x^{k-14})$$

But we're only interested in the exponents of $3$ and $-13$:

  • $k - 15 = 3 \implies k = 18$ (out of bounds)
  • $k - 14 = 3 \implies k = 17$
  • $k - 15 = -13 \implies k = 2$
  • $k - 14 = -13 \implies k = 1$

So, considering only the three terms $k = 1, 2, 17$:

$$f(x) = {17 \choose 1}(x^{-14}-x^{-13}) + {17 \choose 2}(x^{-13}-x^{-12}) + {17 \choose 17}(x^{2}-x^{3}) + \dots$$ $$f(x) = - {17 \choose 1}x^{-13} + {17 \choose 2}x^{-13} - {17 \choose 17}x^{3} + \dots$$ $$f(x) = -17x^{-13} + 136x^{-13} - 1x^{3} + \dots$$ $$f(x) = 119x^{-13} - 1x^{3} + \dots$$

So the sum of the two relevant coefficients is $118$.

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Your reasoning and calculations are correct. This is merely an addition for convenient notation. We can use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance \begin{align*} [x^k](1+x)^n = \binom{n}{k} \end{align*} We already know that \begin{align*} (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5=x^{-15}(1-x)(1+x)^{17} \end{align*} We obtain \begin{align*} {\color{blue}{[x^3]}}{\color{blue}{x^{-15}(1-x)(1+x)^{17}}} &=[x^{18}](1-x)(1+x)^{17}\\ &=\left([x^{18}]-[x^{17}]\right)(1+x)^{17}\\ &\,\,\color{blue}{=-1} \end{align*} and \begin{align*} {\color{blue}{[x^{-13}]}}{\color{blue}{x^{-15}(1-x)(1+x)^{17}}} &=[x^{2}](1-x)(1+x)^{17}\\ &=\left([x^2]-[x^1]\right)(1+x)^{17}\\ &\,\,\color{blue}{=\binom{17}{2}-\binom{17}{1}} \end{align*} in accordance with your result.

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