Let $$a_n=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor}\binom{n-k}{k}$$

Then, $a_1=a_2=1$. It can also be shown that $a_n=a_{n-1}+a_{n-2}$ for $n\ge 2$. If so, $\{a_n\}$ will be a Fibonacci sequence which means the desired answer is $$a_n=\boxed{\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right)}$$

(Solving a linear regression, such as Fibonacci sequence, is discussed in the book Competition Algebra.)

The linear regression, $a_n=a_{n-1}+a_{n-2}$, can be derived using casework depending on the parity of $n$.

When $n$ is even, let $n=2m$. Then $$\left\lfloor{\frac{n}{2}}\right\rfloor=m\quad\text{and}\quad\left\lfloor{\frac{n-1}{2}}\right\rfloor=\left\lfloor{\frac{n-2}{2}}\right\rfloor=m-1$$

and $$\begin{align*} a_n=\ &\displaystyle\sum_{k=0}^{m}\binom{n-k}{k}\\=\ &\binom{n}{0} + \sum_{k=1}^{m-1}\binom{n-k}{k} +\binom{n-m}{m} \\=\ &\binom{n-1}{0} + \sum_{k=1}^{m-1}\binom{n-k-1}{k} + \sum_{k=1}^{m-1}\binom{n-k-1}{k-1}+\binom{m}{m} \\ =\ & \sum_{k=0}^{m-1}\binom{n-k-1}{k} + \sum_{k=0}^{m-2}\binom{n-k-2}{k}+\binom{m-1}{m-1} \\=\ & a_{n-1} + \sum_{k=0}^{m-1}\binom{n-k-2}{k} \\=\ & a_{n-1}+a_{n-2}\end{align*}$$

When $n$ is odd, the regression can also be derived in a similar way.