Summation Equation, Trying to solve this recurrence forumla.

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.

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.:

[tex]\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)[/tex]

where

[tex]f(x) = \sum_{n=1}^{\infty}x^n = \sum_{n=0}^{\infty}x^n - 1 = \frac{1}{1-x} - 1 = \frac{x}{1-x}[/tex]

So

[tex]f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}[/tex]

So

[tex]f'(a) = \frac{1}{(1-a)^2}[/tex]

Finally, the desired number is:

[tex]a(1-a)\frac{1}{(1-a)^2} = \frac{a}{1-a}[/tex]

bobbybob

thank you very much