how to reduce $(1-\alpha)^{T-i}$ into a sum

Question

I'm given the following proof:

\begin{align} & \sum^T_{i=k} \alpha^i(1-\alpha)^{T-i} \binom{T}{i} \\&=\sum^T_{i=k} \alpha^i \binom{T}{i} \sum_{j=0}^{T-i}(-\alpha)^j\binom{T-i}{j} \\&=\alpha^k \sum^{T-k}_{i=0} \frac{T!i!}{(T-k)!(i+k)!} \sum^{T-k-i}_{j=0} \alpha^i (-\alpha)^j \frac{(T-k)!}{i!j!(T-k-i-j)!} \end{align}

I would like to know how $$(1-\alpha)^{T-i}$$ turned into $$\sum_{j=0}^{T-i}(-\alpha)^j\binom{T-i}{j}$$ and the rules used to turn second line to third line.

Answer 1

Use the binomial theorem

$$ (x + y)^n = \sum_{k=0}^n {n\choose k} x^{n-k} y^{k} $$

In your case just make $x = 1$, $y = -\alpha$ and $n = T-i$:

$$ (1 -\alpha)^{T-i} = \sum_{k=0}^{T-i} {T-i\choose k} (-\alpha)^{k} $$