# 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:

1. how is the first expression equivalent to the second?

2. how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?

Thanks a lot!

Note that \begin{align} \sum_{k=0}^{K-1}\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k} &=\sum_{k=1}^{K}\binom{K}{k}x^{k}(1-x)^{K-k}\\ &=\left(\sum_{k=0}^{K}\binom{K}{k}x^{k}(1-x)^{K-k}\right)-(1-x)^K\\ &=(x+(1-x))^K-(1-x)^K\\ &=1^K-(1-x)^K \end{align}

• thanks robjohn! So what is that thing returned by Mathematica then? – balllib Oct 27 '11 at 16:02
• $K$ seems to be reserved (note its color is black, not blue). When using $K1$, Mathematica gives $\left(\left(\frac{1}{1-x}\right)^{\text{K1}}-1\right) (1-x)^{\text{K1}}$. – robjohn Oct 27 '11 at 16:44
• thanks! but i still get some output like before, this is what I input in Mathematica: $\text{Simplify}\left[\sum _{k=0}^{n-1} \left( \begin{array}{c} n \\ k+1 \end{array} \right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(n-1-k)\right]$, changing $K$ to $n$. – balllib Oct 27 '11 at 18:57

Simplify[PowerExpand[Simplify[Sum[Binomial[K, k + 1]*x^(k + 1)*(1 - x)^(K - k - 1), {k, 0, K - 1}], K > 0]]] works nicely. The key is in the use of the second argument of Simplify[] to add assumptions about a variable. and using PowerExpand[] to distribute powers.

• thanks J.M. I've tried yours in Mathematica and indeed it gives nice result. However, as soon as I replace your style of the expression with symbolic one, the problem persists. I used the following: $\text{Simplify}\left[\text{PowerExpand}\left[\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left( \begin{array}{c} K \\ k+1 \end{array} \right)*(x{}^{\wedge}(k+1))*(1-x){}^{\wedge}(K-1-k)\right),K>0\right]\right]\right]$. Do you know why? – balllib Oct 27 '11 at 19:04
• If I really want things to work in Mathematica, I often forgo the fancy formatting. Why do you need the fancy formatting here? – J. M. is a poor mathematician Oct 27 '11 at 23:39

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

• thanks for pointing out. i see that in robjohn's answer too. – balllib Oct 27 '11 at 18:01
• wish I could up-vote this but apparently i need 15 points of reputation. still much appreciated. – balllib Oct 27 '11 at 18:02

The following answer makes sense only with some background in probability. Suppose first that $0 \le x\le 1$.

A possibly biased coin has probability $x$ of landing "heads." Toss the coin $K$ times. We compute the probability $P$ of one or more heads in two different ways.

The probability we get exactly $k+1$ heads is $$\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k}.$$ Add up, from $k=0$ to $k=K-1$. We get the probability of $1$ head, plus the probability of $2$ heads, plus the probability of $3$ heads, and so on, up to the probability of $K$ heads. Thus $$P=\sum_{k=0}^{K-1}\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k}.$$

The probability of $0$ heads (all tails) is $(1-x)^K$, so the probability of one or more heads is $1-(1-x)^K$, and therefore $$P=1-(1-x)^K.$$

The argument so far only proves the desired result for $0\le x\le 1$. But note that for any $K$, each of the expressions we are trying to prove equal is a polynomial of degree $K$. But if two polynomials $A(x)$ and $B(x)$ with real coefficients, each of degree $\le K$, are equal at $K+1$ values of $x$, then they are identically equal.