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Registered
Join Date: Sep 2004
Location: Sandton, South Africa
Posts: 916
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Working backwards gives you x (or 0+1x); 2+2x; 6+4x; 14+8x etc.
From this follows that you have a series that can be expressed as follows:
lim (n=7) of [2^(n+1)]+x(2^n)-2 where x=1
Solves for 382
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Last edited by Willem Fick; 02-10-2009 at 07:29 AM..
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