i need help to solve this exercise
To prove the continued fraction of (d^2 +1)^(1/2) is [d;2d,2d,2d,….]
Hello, user!
Prove that the continued fraction of .$\displaystyle \sqrt{d^2 +1}$ . is .$\displaystyle \bigg[d,2d,2d,2d, \hdots\bigg]$
$\displaystyle \text{Let: }\;x \;=\;d + \frac{1}{2d + \dfrac{1}{2d + \hdots}} $
$\displaystyle \text{Add }d\text{ to both sides: }\;x + d \;=\;2d + \frac{1}{\left\{2d + \dfrac{1}{2d + \hdots}\right\}} \;\begin{array}{c} \\ \\ \Leftarrow\text{ This is }(x+d) \end{array}$
$\displaystyle \text{Then we have: }\;x + d \;=\;2d + \frac{1}{x+d}$
$\displaystyle \text{Multiply by }(x+d)\!:\;\;(x+d)^2 \;=\;2d(x+d) + 1 \quad\Rightarrow\quad x^2 + 2dx + d^2 \;=\;2dx + 2d^2 + 1$
$\displaystyle \text{Therefore: }\;x^2 \;=\;d^2+1 \quad\Rightarrow\quad x \;=\;\sqrt{d^2+1}$