How can (eqn.1) be simplified to (eqn.2) using factorials and summation?

babyrudin
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Hello all! In solving some math problems, I encountered the following sum:

\sum_{k=1}^{r+1} kb \frac{r!}{(r-k+1)!} \frac{(b+r-k)!}{(b+r)!}. \quad \mbox{(eqn.1)}

Now, I have asked Maple to calculate the above sum for me, and the answer takes a very simple form:

\frac{b+r+1}{b+1}. \quad \mbox{(eqn.2)}

My question is, does anyone know how to go from (eqn.1) to (eqn.2)? I am really bad at working with factorials, and so far I am not getting close. Maybe there are some results and properties of factorials and summation that can be used to simplify the above?
 
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I suggest assuming specific values for b, r, k.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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