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April 29th, 2014, 12:59 PM   #1
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The sum of the first n terms of the sequence
\[ 1,~(1+2),~(1+2+2^2),~\dots ~(1+2+2^2+\dots+2^{n-1}) \]
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April 29th, 2014, 01:42 PM   #2
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$\displaystyle \sum_{j=0}^{n-1}\sum_{i=0}^{j}2^i = \sum_{j=0}^{n-1}\left(2^{j+1}-1\right) = 2^{n+1}-2-n$
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