Is there a way to simplify $\sum_{n=0}^L \binom{L}{n} e^{p^n q^{L-n}}$? I have a nasty sum I've been working with that can be reduced to the following form
$$ \sum_{n=0}^L \binom{L}{n} e^{-c p^n q^{L-n}}$$
where $c > 0$ is constant. However, I do not know any way to simplify/reduce this. Is there any nice way to do it, or do I need to start approximating the sum? Doing a bit of research suggests no, but I still have  hope for it.
EDIT: Forgot to mention $p+q =1, 0<p<1.$
 A: It's not much, but we can write
\begin{align}
\sum_{n=0}^L \binom{L}{n} \exp\Big(-c \left(p^n q^{L-n}\right)\Big)
&=
\sum_{n=0}^L\, \binom{L}{n}\, \sum_{k\geq 0}\frac{{\left(-c \left(p^n q^{L-n}\right)\right)}^k}{k!}
\\&=
\sum_{k\geq 0}\,(-1)^k\,\frac{c^k}{k!} \, \left(\sum_{n=0}^L\, \binom{L}n\,{\left(p^k\right)}^n{\left(q^k\right)}^{L-n}\right)
\\&=
\sum_{k\geq 0}(-1)^k\,\frac{c^k\,{\left(p^k+q^k\right)}^L}{k!}.
\end{align}
This looks somewhat easier to make estimates on, or else might point someone in a better direction.

We can also attempt some calculus.
\begin{align}
f(p, L, c)
&= \sum_{k\geq 0}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^k}{k!}
\\&=
2^L-c+
\sum_{k\geq 2}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^k}{k!}
\end{align}
Then $\frac{\partial f}{\partial p}(p,L,c)$ equals
\begin{align}
&\hphantom{=}
\sum_{k\geq 2}\,L\,{\left(p^k+(1-p)^k\right)}^{L-1}\left(kp^{k-1}-k(1-p)^{k-1}\right)\,\frac{(-c)^k}{k!}
\\&=
\sum_{k\geq 1}\,L\,{\left(p^k+(1-p)^k\right)}^{L-1}\left(kp^{k-1}-k(1-p)^{k-1}\right)\,\frac{(-c)^k}{k!}
\\&=
-Lc\left[\left(\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^{L-1}\,\frac{(-pc)^{k-1}}{(k-1)!}\right)
-
\left(\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^{L-1}\,\frac{((1-p)c)^{k-1}}{(k-1)!}\right)\right]
\end{align}
On the other hand,
\begin{align}
\frac{\partial f}{\partial c}(p,L, c)
&=
-1-\sum_{k\geq 2}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^{k-1}}{(k-1)!}
\\&=
-\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^{k-1}}{(k-1)!}
\end{align}
Hence, for $L>1$ we have
\begin{align}
\frac{\partial f}{\partial p}(p,L,c)
&=
-Lc\left[
\left(-\frac{\partial f}{\partial c}(p,L-1,pc)\right)
-
\left(-\frac{\partial f}{\partial c}(p,L-1,(1-p)c)\right)
\right]
\\&=
Lc\left[
\frac{\partial f}{\partial c}(p,L-1,pc)
-
\frac{\partial f}{\partial c}(p,L-1,(1-p)c)\right]
\end{align}
While this does not look like a simple DE to solve, it at least shows that $p=\frac12$ is a critical point for fixed $L>1$, $c>0$.
