(New page: Spencer, I am skeptical about your formula. you posted: <math>\sum_{i=1}^n\frac{n}{n - i + 1}\!</math> being equal to <math>\sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}\!</math> Try n=...) |
(No difference)
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Revision as of 17:18, 6 October 2008
Spencer, I am skeptical about your formula.
you posted:
$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $
being equal to
$ \sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}\! $
Try n=3
original expected value would be 11/2 = 5 1/2
your formula yields 6.
My calculation my be wrong, but otherwise I am thinking I might leave the formula as the answer.
