I need to be able to show that: $\sum_{k=i}^{n} {n \choose k} (1-t)^{n-k} t^{k-i} {k \choose i} (1-\tau)^{k-i}$ is equivalent to ${n \choose i} (1-\tau t)^{n-i}$. However I have no idea how to expand this sum out.
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Sorry the binomial coefficient should have been $i$ instead of $k$. I've edited it now. – CasulOrnstein Dec 23 '15 at 13:38
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Let $n=m+i$ and $k=i+j$. Then: $$\begin{eqnarray*}\sum_{k=i}^{n}\binom{n}{k}\binom{k}{i}(1-t)^{n-k}t^{k-i}(1-\tau)^{k-i}&=&\sum_{j=0}^{m}\frac{(m+i)!}{i!j!(m-j)!}(1-t)^{m-j}t^{j}(1-\tau)^{j}\\&=&\binom{n}{i}\sum_{j=0}^{m}\binom{m}{j}(1-t)^{m-j} t^j (1-\tau)^{j}\\&=&\binom{n}{i}\left(1-t+t(1-\tau)\right)^{m}\\&=&\binom{n}{i}(1-t\tau)^{n-i}.\end{eqnarray*}$$
Jack D'Aurizio
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