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Now take the derivative with respect to r and you get:
 
Now take the derivative with respect to r and you get:
  
<math>\sum_{n=0}^\infty n*r^n = 1/(1-r)^2</math>
+
<math>\sum_{n=0}^\infty nr^(n-1) = 1/(1-r)^2</math>
  
 
You can use this equation to simplify your expected value.
 
You can use this equation to simplify your expected value.

Revision as of 11:24, 18 September 2008

The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated for of the commonly used geometric series. $ \sum_{n=0}^\infty r^n = 1/(1-r) $

Now take the derivative with respect to r and you get:

$ \sum_{n=0}^\infty nr^(n-1) = 1/(1-r)^2 $

You can use this equation to simplify your expected value.

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