Simplify $\sum_{i=0}^{N/2}{N-i\choose i}$

It’s well known that

$$\sum_{k\ge 0}\binom{n-k}k=F_{n+1}\;,$$

the $n$-th Fibonacci number, where $F_0=0$, $F_1=1$, and $F_n=F_{n-1}+F_{n-2}$ for $n\ge 2$. This is easily proved by induction on $n$. The induction step is

$$\begin{align*} \sum_{k\ge 0}\binom{n+1-k}k&=\sum_{k\ge 0}\left(\binom{n-k}k+\binom{n-k}{k-1}\right)\\ &=\sum_{k\ge 0}\binom{n-k}k+\sum_{k\ge 0}\binom{n-k}{k-1}\\ &\overset{*}=F_{n-1}+\sum_{k\ge 1}\binom{n-k}{k-1}\\ &=F_{n-1}+\sum_{k\ge 0}\binom{n-1-k}k\\ &\overset{*}=F_{n-1}+F_{n-2}\\ &=F_n\;, \end{align*}$$

where the two starred steps are applications of the induction hypothesis.

Note that there’s no need to specify an upper limit on the summation: once $k$ exceeds $n-k$, the binomial coefficient is $0$ anyway.