Continued Fraction Algorithm for 113/50 The numbers $a_k$ can be found for $\frac{113}{50}$ by using a continued fraction algorithm. Note that $\frac{113}{50}$ is rational, and as a result it will have to terminate. Can anyone help me determine the numbers $a_k$ for $\frac{113}{50}$?
What I have Done:
$\frac{113}{50} = 2.26$ 
$i=2$ 
$2.26 - 2 = .26$ 
$\frac{26}{100}$  
$\frac{100}{26} = 3+ \frac{22}{26} $ 
3 is  $a_1$  
$\frac{26}{22} = 1 + \frac{4}{22}  $
1 is  $a_2$ 
$\frac{22}{4} = 5 + \frac{2}{4}$  
$a_3 = 5$  
$\frac{4}{2} = 2 $ 
$a_4 = 2 $
Is this correct, and what do I do next?

$$\begin{gathered}
  \frac{{113}}{{50}} = \frac{{100}}{{50}} + \frac{{13}}{{50}} = 2 + \frac{1}{{\frac{{39}}{{13}} + \frac{{11}}{{13}}}} = 2 + \frac{1}{{3 + \frac{{11}}{{13}}}} \hfill \\
   = 2 + \frac{1}{{3 + \frac{{11}}{{11 + 2}}}} = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{2}{{10 + 1}}}}}} \hfill \\
   = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{1}{{5 + \frac{1}{2}}}}}}} = [2;3,1,5,2] \hfill \\ 
\end{gathered}$$
