Simplify an expression to show equivalence I am trying to simplify the following expression I have encountered in a book
$\sum_{k=0}^{K-1}\left(\begin{array}{c}
K\\
k+1
\end{array}\right)x^{k+1}(1-x)^{K-1-k}$
and according to the book, it can be simplified to this:
$1-(1-x)^{K}$
I wonder how is it done? I've tried to use Mathematica (to which I am new) to verify, by using
$\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left(
\begin{array}{c}
 K \\
 k+1
\end{array}
\right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(K-1-k)\right)\right]$
and Mathematica returns
$\left\{\left\{-\frac{K q \left((1-q)^K-q^K\right)}{-1+2 q}\right\},\left\{-\frac{q \left(-(1-q)^K+(1-q)^K q+(1+K) q^K-(1+2 K) q^{1+K}\right)}{(1-2 q)^2}\right\}\right\}$
which I cannot quite make sense of it.
To sum up, my question is two-part:


*

*how is the first expression equivalent to the second?

*how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?
Thanks a lot!

It follows from the Binomial Formula, http://en.wikipedia.org/wiki/Binomial_theorem.
