2.2. For any positive integer $n$, evaluate
$$
\sum_{i=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\binom{n-i+1}{i},
$$
where $\binom{m}{k}=\frac{m!}{k!(m-k)!}$ and $\left\lfloor\frac{n+1}{2}\right\rfloor$ is the greatest integer less than or equal to $\frac{n+1}{2}$.
2.2. $\binom{n-i+1}{i}$ is equal to the number of $i$-subsets of the set $S=\{1,2, \ldots, n\}$ containing no consecutive integers. Hence the required sum is just the number $a_{n}$ of subsets of $S$ containing no consecutive integers. It can be shown easily that $a_{n}$ satisfies the recurrence relation: $a_{n}=a_{n-1}+a_{n-2}$ with $a_{0}=1$ and $a_{1}=2$. This can also be derived from the identity:
$$
\binom{n-i+1}{i}=\binom{(n-1)-i+1}{i}+\binom{(n-2)-(i-1)+1}{i-1} .
$$

From this, we obtain
$$
\sum_{i=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\binom{n-i+1}{i}=\frac{5+3 \sqrt{5}}{10}\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\frac{5-3 \sqrt{5}}{10}\left(\frac{1-\sqrt{5}}{2}\right)^{n} .
$$