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I have the following recurrence relationship for which I know the solution is given by a binomial distribution:

reccurance =  A (i + 1) v[i + 1] + (n - i - 1) v[i - 1] == (n - i) v[i] + A i v[i]

But when I use RSolve I get

RSolve[{reccurance}, v, i]

{{v -> DifferenceRoot[
    Function[{\[FormalY], \[FormalN]}, {(-2 - \[FormalN] + 
           n) \[FormalY][\[FormalN]] + (1 + \[FormalN] (1 - A) - A - 
           n) \[FormalY][
          1 + \[FormalN]] + (2 A + \[FormalN] A) \[FormalY][
          2 + \[FormalN]] == 0, \[FormalY][0] == 
       C[1], \[FormalY][1] == C[2]}]]}}

Putting in the appropriate boundary conditions does not help things at all. e.g.

RSolve[{reccurance, v[-1] == 0, v[n + 1] == 0}, v, i]

Does not do anything at all (well, it returns the input). Nor does adding the constraint for all values to sum to one, or the first value equal to a yet-to-be-determined constant.

How do I make this result in something that is clearly a binomial distribution?

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1  
What is $n$? You seem to have two recurrence variables, $i$ and $n$. –  bill s Nov 30 '12 at 13:26
    
@bills n is a integer constant –  Lucas Nov 30 '12 at 17:30
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