The generating function for the Fibonacci numbers Prove that $$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}$$
The coefficients are Fibonacci numbers, i.e., the sequence $\left\{1,1,2,3,5,8,13,21,...\right\}$. 

Since Fibbonacci series starts at $0$ this series is sometimes called shifted Fibonacci sequence
$F_0=1,F_1=1,F_2=2,F_3=3,F_4=5,...$
$$ F_n=F_{n-1}+F_{n-2},n\ge 2$$
Denote by $$F(x)=\sum_{n=0}^{\infty}F_nx^n=1+x+\sum_{n=2}^{\infty}F_nx^n=1+x+\sum_{n=2}^{\infty}(F_{n-1}+F_{n-2})x^n=$$ 
$$=1+x+x\sum_{n=2}^{\infty}F_{n-1}x^{n-1}+x^2\sum_{n=2}^{\infty}F_{n-2}x^{n-2}=$$ 
$$=1+x+x(-1+F_0x^0+\sum_{n=2}^{\infty}F_{n-1}x^{n-1})+x^2\sum_{n=2}^{\infty}F_{n-2}x^{n-2}=$$ 
$$=1+x+x(-1+\sum_{n=1}^{\infty}F_{n-1}x^{n-1})+x^2\sum_{n=2}^{\infty}F_{n-2}x^{n-2}=$$
$$=1+x+x(-1+F(x))+x^2F(x)=1+xF(x)+x^2F(x)$$ 
or
$$F(x)=1+xF(x)+x^2F(x)$$
solving this equation we get
 $$F(x)=\frac{1}{1-x-x^2}$$
