# Summation Equation, Trying to solve this recurrence forumla.

by bobbybob
Tags: equation, forumla, recurrence, solve, summation
 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 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: $$a(1-a)\frac{1}{(1-a)^2} = \frac{a}{1-a}$$