I am attempting to prove a non-trivial upper bound on the following expression.
Let $0 < r \leq 1$, and let $p$ be a positive integer.
My summation is the following:
$\sum_{k=0}^\left\lfloor \frac{p}{2} \right\rfloor {p \choose 2k}r^{2k} = 1 + {p \choose 2}r^2 + {p \choose 4}r^4 + \ldots$
Note that when $r = 1$, I think that it is easy to see that this summation is $2^{p-1}$, as we are essentially counting the number of even subsets of a set of size $p$ (right?).
I'm not sure how to bound it as a function of $r$ and $p$, however.
Any help is greatly appreciated.