Proof that sum of combinatorics and fractions equals $1$

Question

For one of the questions in my statistics textbook, the last step of the solution is

$$ {m \choose x} (pr)^x (1 - rp)^{m-x} \sum_{t=0}^{k} {k \choose t} \bigg( \frac{r - rp}{1 - rp} \bigg)^t \bigg( \frac{1 - r}{1 - rp} \bigg)^{k-t} $$

$$ = {m \choose x} (pr)^x (1 - rp)^{m-x} $$

which implies that

$$ \sum_{t=0}^{k} {k \choose t} \bigg( \frac{r - rp}{1 - rp} \bigg)^t \bigg( \frac{1 - r}{1 - rp} \bigg)^{k-t} = 1 .$$

I know that $\sum_{t=0}^{k} {k \choose t} = 2^k$ but other than that I'm not sure how to proceed.

Answer 1

Binomial expansion: $$(a + b)^n = \sum_{t=0}^n {n \choose t} a^t b^{n-t}$$ Notice in your case $a + b =1$.

Answer 2

For ease of perception, set $a = r-rp$ and $b = 1-r$. Then $a+b=1-rp$. In these terms, the sum is:

$$\begin{align} \sum_{t=0}^{k} {k \choose t} \left( \frac{r - rp}{1 - rp} \right)^t \left( \frac{1 - r}{1 - rp} \right)^{k-t} &= \sum_{t=0}^{k} {k \choose t} \left( \frac{a}{a+b} \right)^t \left( \frac{b}{a+b} \right)^{k-t} \\[1ex] &=\left( \frac{a}{a+b} + \frac{b}{a+b}\right)^k \\[1ex] &=1^k = 1 \end{align}$$