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<math>\sum_{n = M}^N \alpha^n = \alpha^M - \alpha^{N-1}</math>
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<math>\sum_{n = M}^N \alpha^n = \alpha^M - \alpha^{N-1} / (1 - \alpha)</math>

Revision as of 01:42, 20 July 2009

$ \sum_{n = M}^N \alpha^n = \alpha^M - \alpha^{N-1} / (1 - \alpha) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood