Question

Show that:
(1) $\sqrt{a^2+b}=a+\frac{b}{2a+}\frac{b}{2a+}\cdots$,

(2) $\sqrt{a^2-b}=a-\frac{b}{2a-}\frac{b}{2a-}\cdots$.

Answer 1

$\left(1\right)\sqrt{a^2+b}=a+(\sqrt{a^2+b}-a)$

$=a+\frac{b}{\sqrt{a^2+b}+a}$

$\sqrt{a^2+b}+a=2a+(\sqrt{a^2+b}-a)$

$=2a+\frac{b}{\sqrt{a^2+b}+a}$

and so on

$\therefore \sqrt{a^2+b}=a+\frac{b}{2a+}\frac{b}{2a+}...$

$\left(2\right)\sqrt{a^2-b}=a-(a-\sqrt{a^2-b})=a-\frac{b}{a+\sqrt{a^2-b}}$

$a+\sqrt{a^2-b}=2a-(a-\sqrt{a^2-b})=2a-\frac{b}{a+\sqrt{a^2-b}}$

and so on

$\therefore \sqrt{a^2-b}=a-\frac{b}{2a-}\frac{b}{2a-}...$

Therefore, hence showed.