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Define the sequence \(\{x_i\}_{i\ge 0}\) by \(x_0=2009\) and \(x_n=-\frac{2009}{n}\sum_{k=0}^{n-1}x_k\) for all \(n\ge 1\). Compute the value of \(\sum_{n=0}^{2009}2^nx_n\).

 May 8, 2019
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The answer is 2010.

 Nov 29, 2019

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