The problem:
Show that, as $n\rightarrow + \infty$, $\sum_{k=1}^{n/2} \dbinom{n}{k}\alpha^{k(n-k)} \rightarrow 0$ for $0<\alpha<1$.
What I tried :
I tried to use $\dbinom{n}{k} \leq \dbinom{n}{\lfloor{n/2}\rfloor}$ After that : $$ \begin{align} \sum_{k=1}^{n/2} \dbinom{n}{k}\alpha^{k(n-k)}& \leq \dbinom{n}{\lfloor{n/2}\rfloor}\sum_{k=1}^{n/2}\alpha^{k(n-k)} \\ & \leq \dbinom{n}{\lfloor{n/2}\rfloor}\sum_{k=1}^{n/2}\alpha^k && \text{(cause } \alpha<1) \\ & \leq \dbinom{n}{\lfloor{n/2}\rfloor}\dfrac{1-\alpha^{\lfloor n/2\rfloor}}{1-\alpha} \end{align}$$ but it is not good enough because $\dbinom{n}{\lfloor{n/2}\rfloor}\dfrac{1-\alpha^{\lfloor n/2\rfloor}}{1-\alpha} \rightarrow +\infty$
Fix $1\leq k_0\leq n/2$ so that $\alpha^{k_0} < 1/2$ and split the sum as $$ \sum_{k=1}^{n/2} \dbinom{n}{k} \alpha^{k(n-k)} = \sum_{k=1}^{k_0} + \sum_{k = k_0+1}^{n/2} := A + B, $$ we will estimate $A$ and $B$ separately.
First observe that $$ A \leq \alpha^{n-1} \sum_{k=1}^{k_0} \dbinom{n}{k} \leq \alpha^{n-1} n^{k_0}, $$ where we used a crude estimate $\dbinom{n}{k} = \frac{n(n-1)...(n-k_0 +1)}{k_0!} \leq n^{k_0}$. Since $k_0$ is fixed and $\alpha<1$ we get $$ A \to 0 \text{ as } n \to \infty. $$
We now proceed to the second part $B$. Consider the function $f(x) = x (n-x)$, where $x\geq 1$. Since $f'(x) = n - 2x $, we see that $f$ increases in the range $1\leq x \leq n/2$. Hence we have $k(n-k) \geq k_0(n-k_0)$ for all $k=k_0+1,...,n/2$, from which we obtain $$ B \leq \sum_{k= k_0 + 1}^{n/2} \dbinom{n}{k} \alpha^{k_0(n-k_0)} \leq \alpha^{k_0(n-k_0)} \sum_{k=0}^n \dbinom{n}{k} = \alpha^{k_0(n-k_0)} 2^n = (2\alpha^{k_0})^n \alpha^{-k_0^2}. $$ But recall, that $k_0$ was fixed so that $2\alpha^{k_0}<1$, and hence $B \to 0$ as $n\to \infty$.
We thus proved that both $A$ and $B$ converge to $0$, hence so does the original sum.