Prove that ${F_{n+1}} = \sum_{k=0}^{n}\binom{n-k}{k}$ where ${F_{n}}$ is n-th Fibonacci number

Question

Question:

Prove that ${F_{n+1}}$ = $\sum_{k=0}^{n} \binom{n-k}{k}$ where ${F_{n}}$ is n-th Fibonacci number.

My idea:

I guess I could use induction so ${F_{n}}\to{F_{n+1}}$ where we see that for $n=0$ and $n=1$ it is true so that would be our base. Then - ${F_{n+1}}= \sum_{k=0}^{n+1} \binom{n-k+1}{k}$ but I have no idea what to do with it, does anyone have any idea? Maybe use some binomial or Fibonacci properties?

How does that help me though? I have two arguments at the top of the binomial. — Michał, Apr 21 '18 at 19:33

score 0 · Accepted Answer · answered Apr 21 '18 at 19:45

Hint: $$ F_{n+1}=F_{n}+F_{n-1}\stackrel{I.H.}=\sum_{k=0}^{n-1} \binom{n-k-1}{k}+\sum_{k=0}^{n-2} \binom{n-k-2}{k}\\=\sum_{k=0}^{n-1} \binom{n-k-1}{k}+\sum_{k=1}^{n-1} \binom{n-k-1}{k-1}=\sum_{k=0}^{n}\binom{n-k-1}{k}+\binom{n-k-1}{k-1}=\sum_{k=0}^{n}\binom{n-k}{k}. $$

Prove that ${F_{n+1}} = \sum_{k=0}^{n}\binom{n-k}{k}$ where ${F_{n}}$ is n-th Fibonacci number

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