# Summation Equation, Trying to solve this recurrence forumla.

1. Mar 5, 2006

### bobbybob

Hello. I've searched around a bit for a math forum where I could get help with this and this seems like the one I found where I could get some help with this. I was posed the following problem. Now I must admit it is over my head (as is most of the math on this forum) I was hoping that someone here could help give me an answer for this person. Or if not an answer, a reason why his probem makes no sense.

2. Mar 5, 2006

### AKG

You either miss your first shot, OR make your first and miss the next, OR make your first two and miss the next OR ... OR make the first n and miss the next OR ...

The probability that you make the first n and miss the next is an(1-a).

So what you want is the sum of (getting n cups) x (probability of getting n cups) for all n, i.e.:

$$\sum _{n=0}^{\infty}na^n(1-a) = (1-a)\sum _{n=0}^{\infty}na^n = (1-a)\sum _{n=1}^{\infty}na^n = a(1-a)\sum_{n=1}^{\infty}na^{n-1} = a(1-a)f'(a)$$

where

$$f(x) = \sum_{n=1}^{\infty}x^n = \sum_{n=0}^{\infty}x^n - 1 = \frac{1}{1-x} - 1 = \frac{x}{1-x}$$

So

$$f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$$

So

$$f'(a) = \frac{1}{(1-a)^2}$$

Finally, the desired number is:

$$a(1-a)\frac{1}{(1-a)^2} = \frac{a}{1-a}$$

3. Mar 5, 2006

### bobbybob

thank you very much