Stirling's approximation for $ \sum _{k=1}^{n/2} \frac{n!}{k!k!(n-2k)!}a^kb^kc^{n-2k}$ Suppose $a$, $b$, and $c$ are positive real numbers satisfying $a+b+c=1$. I am trying to use Stirling's approximation to obtain an asymptotic (computable) formula for
$$ \sum _{k=1}^{n/2} \frac{n!}{k!k!(n-2k)!}a^kb^kc^{n-2k}$$
as $n \to \infty$. If $n$ is odd then the upper limit of the sum should actually be $(n-1)/2$.
The problem is how to deal with the denominator, seeing as when $n\to \infty$, as far as I understand it, you can't really use asymptotic formulas for $k!$, since $k$, for instance, will take low values for some terms in the summation.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[8px,#ffe]{\ds{\sum _{k = 1}^{n/2}{n! \over k!\,k!\pars{n - 2k}!}\,a^{k}b^{k}c^{n - 2k}}} =
-c^{n} +
\sum _{k = 0}^{\infty}{n! \over k!\,k!\pars{n - 2k}!}\,a^{k}b^{k}c^{n - 2k}
\\[5mm] & =
-c^{n} +
\sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}}
{n! \over k!\,p!\,q!}\,a^{k}b^{p}c^{q}\bracks{k + p + q = n}[p = k]
\\[5mm] & =
-c^{n} + \sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}}
{n \choose k,p,q}\,a^{k}b^{p}c^{q}
\bracks{k + p + q = n}\braces{\vphantom{\Large A}\bracks{z^{0}}z^{p - k}}
\\[5mm] & =
-c^{n} +
\bracks{z^{0}}\sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}}
{n \choose k,p,q}\pars{a \over z}^{k}\pars{bz}^{\,p}c^{q}
\bracks{k + p + q = n}
\\[5mm] & =
-c^{n} + \bracks{z^{0}}\pars{{a \over z} + bz + c}^{n} =
-c^{n} + \bracks{z^{0}}
{\pars{bz^{2} + cz + a}^{n} \over z^{n}}
\\[5mm] & =
-c^{n} + b^{n}
\bracks{z^{n}}\pars{z^{2} + {c \over b}\,z + {a \over b}}^{n} =
\bbx{\ds{-c^{n} + b^{n}
\bracks{z^{n}}\braces{\vphantom{\Large A}\pars{z - r_{-}}^{n}
\pars{z - r_{+}}^{n}}}}
\end{align}
where $\ds{\braces{r_{-},r_{+}}}$ are the roots of
$\ds{z^{2} + {c \over b}\,z + {a \over b} = 0}$. Namely,
$$
r_{\pm} = {-c \pm \,\mrm{sgn}\pars{b}\root{c^{2} - 4a^{2}} \over 2b}
$$

Further progress is available whenever particular values of $\ds{a,b,c}$ are known.

