# need help about sequence of fibonacci

if you cant see or that picture seem not clear , please feel free to ask me, thank you

Note by NurFitri Hartina
4 years, 11 months ago

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The generating function for the Fibonacci numbers is $$G(x) = \frac{x}{1-x-x^2}$$. (Think of the infinite geometric series and what terms combine to make $$x^n$$...). So, you want to solve $$\frac{a}{1-a-a^2} = 2$$ or $$2a^2 +3a -2 = 0$$. Since we must have $$0<a<\frac{\sqrt{5}-1}{2}$$, this leaves $$a=\frac{1}{2}$$.

- 4 years, 11 months ago

Could you elaborate on how you found that closed form?

- 4 years, 11 months ago

Here is an interesting idea. Try to follow these steps carefully:

$$G(x) = x+x^2+2 x^3+3 x^4+5 x^5+8 x^6+13 x^7+\text{...}$$

$$\frac{G(x)}{x} = 1+x+2 x^2+3 x^3+5 x^4+8 x^5+13 x^6+\text{...}$$

$$\frac{G(x)}{x}-G(x) = 1+x^2+x^3+2 x^4+3 x^5+5 x^6+8 x^7+\text{...}$$

$$x G(x) = x^2+x^3+2 x^4+3 x^5+5 x^6+8 x^7+13 x^8+\text{...}$$

And finally,

$$\frac{G(x)}{x}-G(x)-x G(x)=1$$

Solve this for $$G(x)$$ to get $$G(x) = \frac{x}{1−x−x^2}$$

- 4 years, 11 months ago

Wow, that's amazing. It's like the first time I saw a proof of the sum of a geometric series, where suddenly all the terms canceled out.

- 4 years, 11 months ago

I highly recommend Wilf's Generatingfunctionology, which deals with formal generating functions. The second edition is available online for free on the author's site. Using this, you can actually generate a formal proof.

- 4 years, 11 months ago

That's looking really good, thank you.

- 4 years, 11 months ago

that's very nice book, thanks

- 4 years, 11 months ago