TITLE: The coefficient of $x^{101}$ in the expansion of $(1-x)(1-2x)(1-2^2x)...(1-2^{101}x) =?$
QUESTION [6 upvotes]: 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

REPLY [5 votes]: 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*}