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Summation Equation, Trying to solve this recurrence forumla. 
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#1
Mar506, 06:12 PM

P: 2

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
Mar506, 07:01 PM

Sci Advisor
HW Helper
P: 2,586

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 a^{n}(1a). 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(1a) = (1a)\sum _{n=0}^{\infty}na^n = (1a)\sum _{n=1}^{\infty}na^n = a(1a)\sum_{n=1}^{\infty}na^{n1} = a(1a)f'(a)[/tex] where [tex]f(x) = \sum_{n=1}^{\infty}x^n = \sum_{n=0}^{\infty}x^n  1 = \frac{1}{1x}  1 = \frac{x}{1x}[/tex] So [tex]f'(x) = \frac{(1x)  x(1)}{(1x)^2} = \frac{1}{(1x)^2}[/tex] So [tex]f'(a) = \frac{1}{(1a)^2}[/tex] Finally, the desired number is: [tex]a(1a)\frac{1}{(1a)^2} = \frac{a}{1a}[/tex] 


#3
Mar506, 07:18 PM

P: 2

thank you very much



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