Fibonacci numbers proof

Question

Is there any proof that $\sum_{k=0}^{n}$ ${n-k}\choose{k}$ $=f_{n+1}$, when $f_{n}$ is a Fibonacci number? (I tried to write in some $n$'s and it looked true, but I don't really know how should I prove it.

Answer 1

Hint: Let $g_n = \sum_{k=0}^{n}{{n-k}\choose{k}}$. Prove that $g_n = f_{n+1}$ for $n=0,1$ and that $g_{n}=g_{n-1}+g_{n-2}$ for $n\ge 2$. You'll most probably need to use Pascal's rule: $${n \choose k}={n-1 \choose k-1}+{n-1 \choose k}$$

Answer 2

Note that $f_{n+1}$ is the number of ways to write $n$ as an ordered sum of ones and twos (where $f_0=0$, $f_1=1$, $f_2=1$ and so on). Classfy such ordered sums based on, $k$, the number of twos in the sum. If there are $k$ twos in the sum, there are $n-2k$ ones in the sum and the number of such sums is $\binom{n-k}{k}$ since we choose the $k$ positions for the twos among the $n-2k+k=n-k$ lettters in total.