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$F_0=1$, $F_1=1$, $F_n=F_{n-1}+F_{n-2}$. The generating function is $-\frac{1}{x^2+x-1}$. I have to expand it to prove that $F_n=\sum_k\binom{k}{n-k}$. Could you help me please?

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    Do you mean $\sum_k \binom{n-k}{k}$ for the sum?2011-10-19
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    @Mike: Those coincide for suitable ranges of $k$.2011-10-19
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    In fact in a sense the version in the question is nicer in that you can let $k$ run from $0$ to infinity.2011-10-19
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    @anon, joriki: Yes, you both are right.2011-10-19

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Expanding as a geometric series and then using the Binomial Theorem, we obtain $$\sum_{k=0}^\infty (x^2+x)^k=\sum_{k=0}^\infty \sum_{l=0}^k{k\choose l}x^{2l+(k-l)}=\sum_{n=0}^\infty\left(\sum_{k+l=n}{k\choose l}\right)x^n$$ Rewrite the inner sum's index $l=n-k$ so that equating coefficients with $\sum F_n x^n$ gives $$F_n=\sum_{k} {k\choose n-k}.$$