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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This generating function is equivalent to $$\sum_{n}^{\infty}F_n x^n=\dfrac{x}{1-x-x^2}$$

where $F_n$ is a fibonacci number. To show this, I need to simplify the above generating function with binomial coefficients using $$(1-z)^{-\alpha+1}=\sum_{k}^{\infty}{\alpha+k\choose k}z^k$$

I'm familiar with using this conversion, but I can't do it successfully in this case. Please give me a hint for conversion.

EDIT: So, as achille hui suggested, I converted as the following: $$\sum_{s}\sum_{k}{k+s\choose k}x^{s+2k+1}=\sum_{s}x^{s+1}(1-x^2)^{1-s}=x(1-x^2)\dfrac{1}{1-\dfrac{x}{1-x^2}}=\dfrac{x(1-x^2)^2}{1-x-x^2}$$

But the last formula clearly doesn't match with what I expected, which is: $$\sum_{n}^{\infty}F_n x^n=\dfrac{x}{1-x-x^2}$$

Where did I make a mistake?

share|cite|improve this question
Rewrite $n$ as $2k+1+s$, the coefficient $\binom{n-k-1}{k}$ is non-zero only for $s \ge 0$. If you convert your sum to one over $k$ and $s$, everything should be obvious. – achille hui Jun 6 '14 at 23:09
That makes sense! – Math.StackExchange Jun 6 '14 at 23:19
up vote 1 down vote accepted

The generating function of the Fibonacci numbers is as follows. \begin{align} \sum_{n=0}^{\infty} F_{n} t^{n} &= \sum_{n=0}^{\infty} \sum_{k=0}^{[(n-1)/2]} \binom{n-k-1}{k} \ t^{n} \\ &= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{n+k}{k} \ t^{n+2k+1} \\ &= \sum_{n=0}^{\infty} (1-t^{2})^{-n-1} \ t^{n+1} \\ &= \frac{t}{1-t^{2}} \ \sum_{n=0}^{\infty} \left( \frac{t}{1-t^{2}} \right)^{n} \\ &= \frac{t}{1-t-t^{2}}. \end{align}

share|cite|improve this answer
Thanks. I found where I made a mistake from your answer! – Math.StackExchange Jun 6 '14 at 23:43

Your Answer


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.