(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=...) |
|||
| Line 17: | Line 17: | ||
My calculation my be wrong, but otherwise I am thinking I might leave the formula as the answer. | My calculation my be wrong, but otherwise I am thinking I might leave the formula as the answer. | ||
| + | |||
| + | ------- | ||
| + | I agree with Ben. <math>\sum_{i=1}^n\frac{n}{n - i + 1} = \frac{n}{n} + \frac{n}{n-1} + \dots + \frac{n}{3} + \frac{n}{2} + \frac{n}{1}</math> is '''not''' an arithmetic series! | ||
| + | |||
| + | -Brian | ||
Latest revision as of 18:29, 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.
I agree with Ben. $ \sum_{i=1}^n\frac{n}{n - i + 1} = \frac{n}{n} + \frac{n}{n-1} + \dots + \frac{n}{3} + \frac{n}{2} + \frac{n}{1} $ is not an arithmetic series!
-Brian
