Can $\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$ be simplified?

Question

I'm trying to compute a complicated thing, and I end up with terms like:

$$\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$$

$a$ and $b$ are nonnegative integers, $0<p<1$.

I don't see a way to make this simpler. Am I missing something or is this the simplest thing that I'm going to get?

Answer 1

Yes, it can be simplified. Note that for $0\leq a\leq b$, $$\binom{i}{a}\binom{b}{i}=\binom{b}{a}\binom{b-a}{i-a}.$$ Hence $$\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}=p^a\binom{b}{a}\sum_{i=a}^{b}\binom{b-a}{i-a}p^{i-a}.$$ Can you take it from here?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = a}^{b}p^{i}{i \choose a}{b \choose i} & = \sum_{i = 0}^{b}{b \choose i}p^{i}\bracks{z^{a}}\pars{1 + z}^{i} = \bracks{z^{a}}\sum_{i = 0}^{b}{b \choose i}\pars{p + pz}^{i} \\[5mm] & = \bracks{z^{a}}\bracks{1 + \pars{p + pz}}^{\, b} \\[5mm] & = \pars{1 + p}^{b}\bracks{z^{a}}\bracks{1 + {p \over 1 + p}\,z}^{\, b} \\[5mm] & \pars{1 + p}^{b}{b \choose a}\pars{p \over 1 + p}^{a} \\[5mm] & = \bbx{{p^{a} \over \pars{1 + p}^{a - b}}\,{b \choose a}} \\ & \end{align}