Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\frac{n}{2}\rfloor\\\lfloor\frac{n}{2}\rfloor\end{pmatrix} = F_{(n+1)}$$
Example: ${6\choose0} + {5\choose1} + {4\choose2} + {3\choose3} = 13 = F_{7}$ 
 A: Let $G_n=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots $.
Then, we have
$$G_1=1,G_2=2$$
$$G_{n}+G_{n+1}=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots +\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}\cdots $$
$$=\binom{n+1}{0}+\left(\binom n0+\binom{n}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\left(\binom{n-2}{2}+\binom{n-2}{3}\right)+\cdots$$
$$=\binom{n+2}{0}+\binom{n+1}{1}+\binom{n}{2}+\cdots=G_{n+2}$$
So, we know that $G_{n}$ is $n+1$-th Fibonacci number.
A: Using the recursion for Pascal's Triangle, $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, we get
$$
\begin{align}
a_n
&=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\dots\\
&=\binom{n-1}{0}+\binom{n-2}{1}+\binom{n-3}{2}+\dots\\
&+\binom{n-1}{-1}+\binom{n-2}{0}+\binom{n-3}{1}+\dots\\[4pt]
&=a_{n-1}+a_{n-2}
\end{align}
$$
Noting that $a_0=1$ and $a_1=1$, we get $a_n=F_{n+1}$.

If there is concern over the term $\binom{n-1}{-1}$, consider the edge of Pascal's Triangle. Looking at it as $\binom{n}{0}=\binom{n-1}{0}+0$ and $\binom{n}{0}=\binom{n-1}{0}+\binom{n-1}{-1}$, we get $\binom{n-1}{-1}=0$. Furthermore, we have the identity $\binom{n-1}{k-1}=\frac kn\binom{n}{k}$. Set $k=0$ and we get $\binom{n-1}{-1}=0$. However we look at it, we get
$$
\binom{n-1}{-1}=0
$$
A: Another way is using generating functions.  The g.f. of $\displaystyle a_n = \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} $ is
$$ \eqalign{G(x) & =\sum_{n=0}^\infty  \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} x^n\cr  &= \sum_{k=0}^\infty \sum_{n=2k}^\infty {n-k \choose k} x^n\cr
&= \sum_{k=0}^\infty x^{2k} \sum_{m=0}^\infty {m+k \choose k} x^m\cr
&= \sum_{k=0}^\infty \dfrac{x^{2k}}{(1-x)^{k+1}} \cr 
&= \dfrac{1}{1-x}\; \dfrac{1}{1 - x^2/(1-x)}\cr& = \dfrac{1}{1-x-x^2}}$$
The g.f. of the Fibonacci numbers $F_n$ is
$$ F(x) = \sum_{n=0}^\infty F_n x^n = \dfrac{x}{1+x-x^2} $$
with $F_0 = 0$, so the g.f. of $F_{n+1}$ is $F(x)/x = 1/(1+x-x^2) = G(x)$.
A: Use induction and a little cheat!
Consider $F_5=\dbinom50+\dbinom41+\dbinom32=1+4+3=8$
Now $\dbinom{n}{k-1}+\dbinom{n}{k}=\dbinom{n+1}{k}$, so for example $\dbinom41+\dbinom42=\dbinom52$.
So adding these like terms gives us the next Fibonacci number, except for the misplaced leading term.
And we have $\dbinom{n}{0}=1$ for all $n\in\mathbb{N}$ so we shift the leading term outwards, which is a bit like cheating really!
There's loads of pictures at Google.
