The generating function for the Fibonacci numbers

The proof is quite simple. Let's write our sum in a compact format:

$$ 1+z+2z^2+3z^3+5z^4+8z^5+... = \sum_{n=0}^\infty F_nz^n $$ Where $F_n$ is the $n$th Fibonacci number, starting with $F_0=F_1=1$, and $F_{n+2}=F_n+F_{n+1}$. It is from here that we will prove what needs to be proven.

$$\begin{align} (1-z-z^2)\sum_{n=0}^\infty F_nz^n &= \sum_{n=0}^\infty F_nz^n - \sum_{n=0}^\infty F_nz^{n+1} - \sum_{n=0}^\infty F_nz^{n+2}\\ &= \sum_{n=0}^\infty F_nz^n - \sum_{n=1}^\infty F_{n-1}z^n-\sum_{n=2}^\infty F_{n-2}z^n\\ &= F_0 + (F_1-F_0)z + \sum_{n=2}^\infty (F_n-F_{n-1}-F_{n-2})z^n \end{align}$$ Now, $F_1=F_0$ and $F_n=F_{n-1}+F_{n-2}$. Therefore,

$$ (1-z-z^2)\sum_{n=0}^\infty F_nz^n = F_0 = 1 $$ And thus

$$ \sum_{n=0}^\infty F_nz^n = \frac{1}{1-(z+z^2)} $$

$\dfrac{1}{1-(z+z^2)}=1+(z+z^2)+(z+z^2)^2....$ The coefficient of $z^n$ is therefore the number of ways of adding 1s and 2s to get $n$. Also, the number of ways to do this is given by the Fibonacci numbers, proving the result.

A related technique. What you have is the ordinary generating function of Fibonacci numbers. Use the recurrence relation of the Fibonacci numbers

$$ F_{n+2} = F_{n+1} + F_{n} $$

to get the generating function. See here for a related problem.

Added: We will derive the ordinary generating function. Let $g(z)=\sum_{n=0}^{\infty} F_n z^n $, $F_0=F_1=1$, then

$$\sum_{n=0}^{\infty} F_{n+2} z^n = \sum_{n=0}^{\infty} F_{n+1} z^n + \sum_{n=0}^{\infty} F_{n} z^n $$

$$\implies \sum_{n=2}^{\infty} F_{n} z^{n-2} = \sum_{n=1}^{\infty} F_{n} z^{n-1} + g(z) $$

$$\implies \frac{1}{z^2}\sum_{n=2}^{\infty} F_{n} z^{n} =\frac{1}{z} \sum_{n=1}^{\infty} F_{n} z^{n} + g(z) $$

$$ \implies \frac{1}{z^2}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z^2}-\frac{F_1}{z}= \frac{1}{z}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z} + g(z) $$

$$ \implies \frac{g(z)}{z^2}-\frac{1}{z^2}-\frac{1}{z} = \frac{1}{z}g(z)-\frac{1}{z} + g(z) $$

$$ \implies g(z) = \frac{1}{1-(z+z^2)}. $$