Question

Prove that $\sqrt{9a^2+3}=3a+\frac{1}{2a+}\frac{1}{6a+}\frac{1}{2a+}\frac{1}{6a+}....$, and find the fifth convergent.

Answer 1

$\sqrt{9a^2+3}=3a=(\sqrt{9a^2+3}-3a)=3a+\frac{3}{\sqrt{9a^2+3}+3a}$

$\Rightarrow \frac{\sqrt{9a^2+3}+3a}{3}=2a+\frac{\sqrt{9a^2+3}-3a}{3}$

$=2a+\frac{1}{\sqrt{9a^2+3}+3a}$

$\sqrt{9a^2+3a}+3a=6a+\sqrt{9a^2+a}-3a$

$=6a+\frac{3}{\sqrt{9a^2+3}+3a}$

$\therefore \sqrt{9a^2+3}=3a+\frac{1}{2a+}\frac{1}{6a+}\frac{1}{2a+}\frac{1}{6a+}.....$

And the convergents are:-

$\frac{3a}{1},\frac{6a^2+1}{2a},\frac{36a^{2}9a}{12a^2+1},\frac{72a^4+24a^2+1}{24a^3+4a},\frac{432a^5+180a^3+15a}{144a^4+36a^2+1}$