Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$F_0=1$, $F_1=1$, $F_n=F_{n-1}+F_{n-2}$. 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?

share|improve this question
    
Do you mean $\sum_k \binom{n-k}{k}$ for the sum? –  Mike Spivey Oct 19 '11 at 5:26
    
@Mike: Those coincide for suitable ranges of $k$. –  joriki Oct 19 '11 at 5:38
1  
In fact in a sense the version in the question is nicer in that you can let $k$ run from $0$ to infinity. –  joriki Oct 19 '11 at 5:44
    
@anon, joriki: Yes, you both are right. –  Mike Spivey Oct 19 '11 at 6:24

1 Answer 1

up vote 7 down vote accepted

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 $$F_n=\sum_{k} {k\choose n-k}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.