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We have the expression $$\left(\frac{1}{x^2}+\frac{1}{x}+1+x+x^2+ \dots\, +x^{10}\right)^{11}$$ and we have to find the coefficient of $x^{10}$

So I figured out that we need to find coefficient of $x^{32}$ in $\left(\sum_{k=0}^{12}x^k\right)^{11}$. This requires multinomial theorem and explicit calculation would be messy.

Another approach would be to sum the series and find coefficient of $x^{32}$ in $\left(\frac{x^{13}-1}{x-1}\right)^{11}$ And again I get stuck here.

Please help!

MAN-MADE
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akhmeteni
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2 Answers2

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We obtain \begin{align*} \color{blue}{[x^{10}]}&\color{blue}{\left(\frac{1}{x^2}+\frac{1}{x}+1+x+\cdots+x^{10}\right)^{11}}\\ &=[x^{10}]x^{-22}\left(1+x+x^2+\cdots+x^{12}\right)^{11}\\ &=[x^{32}]\left(\frac{1-x^{13}}{1-x}\right)^{11}\\ &=[x^{32}]\left(1-\binom{11}{1}x^{13}+\binom{11}{2}x^{26}\right)\sum_{j=0}^\infty \binom{-11}{j}(-x)^j\tag{1}\\ &=\left([x^{32}]-11[x^{19}]+50[x^{6}]\right)\sum_{j=0}^\infty \binom{10+j}{j}x^j\tag{2}\\ &=\binom{42}{10}-11\binom{29}{10}+55\binom{16}{6}\tag{3}\\ &\color{blue}{=1\,251\,553\,303} \end{align*}

Comment:

  • In (1) we expand the first three terms of the numerator only, since all other terms do not contribute to $[x^{32}]$ and we apply the binomial series expansion.

  • In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and the linearity of the coefficient of operator $[x^k]$.

  • In (3) we select the coefficients of $x^k$ accordingly and apply the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

Markus Scheuer
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  • Thank you! How do we know the second identity? Isn't binomial coefficient defined for positive integers? – akhmeteni Aug 03 '17 at 17:28
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    @akhmeteni: You're welcome! The binomial coefficient $\binom{p}{q}$ is defined for $p\in\mathbb{R}$ and integer values $q$. The reference given in this answer might be helpful. – Markus Scheuer Aug 03 '17 at 17:40
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You need to find coefficient of $x^{32}$ in the expression $$(1+x+x^2+\dots +x^{12})^{11}$$

I don't know the answer will be nice-looking. But this following process may reduce calculation a little.

$(1+x+x^2+\dots +x^{12})^{11}=((1+x+x^2)+x^3(1+x+\dots +x^9))^{11}=\sum_{k=0}^{11}\binom{11}{k}(1+x+x^2)^kx^{33-3k}(1+x+\dots +x^9)^{11-k}$

Note that, first term and last term do not contain $x^{32}$.

Now collect the term which contains $x^{32}$, and continue these process. Similarly apply the same thing on $(1+x+\dots +x^9)^{k}$. I guess this will take a long time to solve.

MAN-MADE
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