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The coefficient of $x^{101}$ in the expansion of $(1-x)(1-2x)(1-2^2x)...(1-2^{101}x) =?$

Can I get some help please?Thanks in advance

humto
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  • Hint: $a_1a_2+a_2a_3+a_3a_1=a_1a_2a_3(a_1^{-1}+a_2^{-1}+a_3^{-1})$ (now generalize to $101$ terms...) – anon Aug 30 '13 at 03:40
  • did not get your cllue – humto Aug 30 '13 at 03:40
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    For example, the $x^2$ coefficient of $(1-a_1x)(1-a_2x)(1-a_3x)$ is $a_1a_2+a_2a_3+a_3a_1$. For future reference sometimes a hint takes more than $30$ seconds to get. – anon Aug 30 '13 at 03:42
  • The Maple code $$a := mul(1-2^n*x, n = 0 .. 101): coeff(a, x^{101}); $$ calculates it. – user64494 Aug 30 '13 at 04:33

1 Answers1

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This is a product of 102 terms, so that the highest-order term in the resulting polynomial is $x^{102}$. The coefficient of $x^{101}$ thus will be

\begin{align*} &\,1\times(-2^1)\times(-2^2)\times\cdots\times(-2^{101})\\ +&\,(-2^0)\times1\times(-2^2)\times\cdots\times(-2^{101})\\ +&\,(-2^0)\times(-2^1)\times1\times\cdots\times(-2^{101})\\ \vdots&\,\\ =&\,(-1)^{101}\sum_{n=0}^{101}\prod_{\substack{m\in\{0,\ldots,101\}\\m\neq n}}2^m\\ =&\,-\sum_{n=0}^{101}2^{(0+\ldots+n-1+n+1+\ldots+101)}\\ =&\,-\sum_{n=0}^{101}2^{(101\times102)/2-n}=-\sum_{n=0}^{101}2^{5\mathord{,}151-n}\\ =&\,-2^{5\mathord{,}151}\sum_{n=0}^{101}2^{-n}=-2^{5\mathord{,}151}(2-2^{-101})=-2^{5\mathord{,}152}+2^{5\mathord{,}050}. \end{align*}

triple_sec
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