Question

In the continued fraction $\frac{b}{a+}\frac{b}{a+}\frac{b}{a+}\cdots ,$
Show that $p_{n+1}=b{q}_n$, $b{q}_{n+1}-a{p}_{n+1}=b^2{q}_{n-1}$.

Answer 1

The series $p=p_1+p_{2}x+p_3{x}^2+....$

$q=q_1+q_{2}x+q_3{x}^2+....$

are both recurring series in which the scale of relation is $1-ax-b{x}^2$

Also, $p_1=b,p_2=ab;q_1=a,q_2=a^2+b;$

$p=\frac{p_1+(p_2-a{p}_1)x}{1-ax-b{x}^2}=\frac{b}{1-ax-b{x}^2}$

$q=\frac{q_1+(q_2-a{q}_1)x}{1-ax-b{x}^2}=\frac{a+bx}{1-ax-b{x}^2}$

$\therefore xq=\frac{ax+b{x}^2}{1-ax-b{x}^2}=\frac{1}{1-ax-b{x}^2}$

Thus $\frac{p_{n+1}}{b}=q_n=$ the co efficient of $x^n$ in $\frac{1}{1-ax-b{x}^2}$

Again $b{q}_{n+1}-a{p}_{n+1}=p_{n+2}-a{p}_{n+1}$

$=b{p}_n=b^2{q}_{n-1}$