# Proof of closed form expression for a sum

For the following sum

$$\sum_{i=0}^{n-r} \binom{n-r}{i}s^{r+i}(1-s)^{n-r-i}(1-t)^{i}$$

Wolfram Alpha finds the following closed form

$$s^{r}(1-st)^{n-r}.$$

However, I have been unable to derive this myself. Without the binomial coefficient, it would be straightforward to simplify along these lines:

$$S = 1 + x + x^2 + x^3 + ... + x^n\\ Sx = x + x^2 + x^3 + ... + x^{n} + x^{n+1}\\ Sx+1 = S + x^{n+1}\\ S=\frac{x^{n+1}-1}{x-1}$$

When $x \neq 1$. But I have not been able to make this approach work since each term ends up with the binomial coefficient of the term that precedes it, thereby preventing the appearance of $S$ on the RHS. Any help is greatly appreciated.

• $$\begin{array}{ll} \sum_{i=0}^{n-r} \binom{n-r}{i}\color{Blue}{s^r}\color{Green}{s^i}(1-s)^{n-r-i}\color{Green}{(1-t)^{i}} & =\color{Blue}{s^r}\sum_{i=0}^{n-r}\color{Green}{(s-st)^i}(1-s)^{n-r-i} \\ & =s^r(\color{Green}{s-st}+1-s)^{n-r}=\cdots\end{array}$$ – anon Jun 26 '13 at 21:04

\begin{align} &\sum_{i=0}^{n-r} \binom{n-r}{i}s^{r+i}(1-s)^{n-r-i}(1-t)^{i}\\ &=s^r(1-s)^{n-r} \sum_{i=0}^{n-r}\binom{n-r}{i}\left(\frac{s(1-t)}{1-s}\right)^i\\ &=s^r(1-s)^{n-r}\left(1+\frac{s(1-t)}{1-s}\right)^{n-r}\\[6pt] &=s^r(1-st)^{n-r} \end{align}
Hint: Write $s^{r+i}(1-t)^i=s^r(s-st)^i$. Factor $s^r$ out and you're left with a standard binomial expansion.