TITLE: How to show that this binomial sum satisfies the Fibonacci relation?
QUESTION [18 upvotes]: The binomial sum
  $$s_n=\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}+\cdots$$
  satisfies the Fibonacci relation.

I failed to prove that $\binom{n-k+1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k}$... Any hints or suggestions?

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}\color{#66f}{\large s_{n}}&
={n+1 \choose 0} + {n \choose 1} + {n - 1 \choose 2} + \cdots
=\sum_{k\ =\ 0}^{\infty}{n + 1 - k \choose k}
\\[5mm]&=\sum_{k\ =\ 0}^{\infty}\ \overbrace{\oint_{\verts{z}\ =\ \varphi^{+}}
{\pars{1 + z}^{n + 1 - k} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ \dsc{n + 1 - k \choose k}}}
\end{align}

where $\ds{\varphi \equiv {1 + \root{5} \over 2}}$ is the
  Golden Ratio.
  Then,

\begin{align}
\color{#66f}{\large s_{n}}&
=\oint_{\verts{z}\ =\ \varphi^{+}}{\pars{1 + z}^{n + 1} \over z}
\sum_{k\ =\ 0}^{\infty}\bracks{1 \over z\pars{1 + z}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ \varphi^{+}}{\pars{1 + z}^{n + 1} \over z}
{1 \over 1 - 1/\bracks{z\pars{1 + z}}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ \varphi^{+}}{\pars{1 + z}^{n + 2} \over z^{2} + z - 1}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ \varphi^{+}}
{\pars{1 + z}^{n + 2} \over \pars{z + \varphi}\pars{z - \varphi^{-1}}}
\,{\dd z \over 2\pi\ic}
={\pars{1 - \varphi}^{n + 2} \over -\varphi - \varphi^{-1}}
+{\pars{1 + \varphi^{-1}}^{n + 2} \over \varphi^{-1} + \varphi}
\end{align}

However,
  $\ds{\quad\varphi + \varphi^{-1}=\root{5}\,,\quad 1 + \varphi^{-1}=\varphi\quad}$ such that

$$
\color{#66f}{\large s_{n}}
={\varphi^{n + 2} - \pars{1 - \varphi}^{n + 2} \over \root{5}}
=\color{#66f}{\large F_{n + 2}}
$$

where $\ds{F_{m}}$ is a
  Fibonacci Number.