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Determine the coefficient of $~x^n~$ in:

$$(x^2 + x^4 + x^6 + ... + x^{n-1})(x + x^3 + x^5 + ... + x^{n-2})$$

Where $~n~$ is an odd number.

How to describe the possible combinations of coefficients that result in $~x^n~$?

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2 Answers 2

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One way is directly, each term left-to-right in the left product must be associated with the corresponding right-to-left term in the right product, giving a total of one product for each element, hence $(n-1)/2$.

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  • $\begingroup$ Is there any more algebraic method? $\endgroup$
    – gmn_1450
    Jun 26, 2019 at 17:12
  • $\begingroup$ @gmn_1450 Certainly, but why use a more difficult method? This is similar to the insight Gauss used to quickly sum the whole numbers from $1$ to $100$. $\endgroup$
    – amd
    Jun 26, 2019 at 19:37
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We use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series.

We consider odd $n=2m+1$ and obtain \begin{align*} \color{blue}{[x^{2m+1}]}&\color{blue}{\left(x^2+x^4+\cdots+x^{2m}\right)\left(x^1+x^3+\cdots+x^{2m-1}\right)}\\ &=[x^{2m+1}]\sum_{j=1}^mx^{2j}\sum_{k=1}^{m}x^{2k-1}\\ &=\sum_{j=1}^m[x^{2m+1-2j}]\sum_{k=1}^mx^{2k-1}\tag{1}\\ &=\sum_{j=1}^m[x^{2j-1}]\sum_{k=1}^mx^{2k-1}\tag{2}\\ &=\sum_{j=1}^m1\tag{3}\\ &\,\,\color{blue}{=m} \end{align*}

Comment:

  • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

  • In (2) we change the order of summation $j \to m-j+1$.

  • In (3) we select the coefficient of $x^{2j-1}$.

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