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Prove the following fibonacci sequence, which appear in Pascal's Triangle. I am not sure where to start on this, any pointers? $$ f_n = {n\choose0} + {n-1\choose1} + ... + {n-k\choose k}$$

where $\displaystyle k=\left\lfloor\frac{n}{2}\right\rfloor$

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up vote 7 down vote accepted

The proof is simpler if you include terms where the lower argument is greater than the upper argument: that is, $$ F_{n+1}=\sum_{k=0}^{n}\binom{n-k}{k}\tag{1} $$ The terms where $n-k\lt k$ are $0$.

Initial Values:

For $n=0$, the sum gives $1$.

For $n=1$, the sum gives $1$.


The recursion is satisfied: $$ \begin{align} \hspace{-1cm}\sum_{k=0}^{n}\binom{n-k}{k}+\sum_{k=0}^{n+1}\binom{n+1-k}{k} &=\sum_{k=1}^{n+1}\binom{n+1-k}{k-1}+\sum_{k=0}^{n+1}\binom{n+1-k}{k}\tag{2}\\ &=1+\sum_{k=1}^{n+1}\binom{n+2-k}{k}\tag{3}\\ &=\sum_{k=0}^{n+2}\binom{n+2-k}{k}\tag{4} \end{align} $$ Explanation:
$(2)$: reindex the left sum $k\mapsto k-1$
$(3)$: pull out the $k=0$ term from the right sum and use $\binom{n+1}{k}=\binom{n\vphantom{1}}{k}+\binom{n\vphantom{1}}{k-1}$
$(4)$: put back the $k=0$ term

Note, however, that because of the initial values, the sum is actually $F_{n+1}$, and not $F_n$.

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Hint: Try to use induction and:

$$ \begin{align} \tag{i} f_{n+2} &= f_{n+1} + f_n \\ \tag{ii} \binom{n}{k} &= \binom{n-1}{k} + \binom{n-1}{k-1} \qquad \text{for $n > 0$} \end{align} $$

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A combinatorial proof.

Fibonacci numbers count the number of ways to walk up $n$ steps, going one or two steps at a time.

$\binom{n-k}{k}$ counts the number of ways of walking up $n$ steps, going one or two steps at a time, but doing exactly $k$ two-steps.

(There are $n-2k$ one-steps, for a total of $n-k$ steps, of which you choose $k$ to be the two-steps)

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Very interesting (+1) – robjohn Jul 8 '14 at 0:59
@robjohn: Yeah it is interesting. In fact, this approach provides many other identities too. – Aryabhata Jul 8 '14 at 1:07

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