Answer
$\approx 2.236$
Work Step by Step
We have to determine an approximation of $\sqrt 5$.
We start with an approximation:
$a_0=2$
Compute $a_1,a_2,a_3,a_4,a_5$ using the formula:
$a_n=\dfrac{1}{2}\left(a_{n-1}+\dfrac{p}{a_{n-1}}\right)$, where $p=5$.
$a_1=\dfrac{1}{2}\left(a_0+\dfrac{5}{a_0}\right)=\dfrac{1}{2}\left(2+\dfrac{5}{2}\right)=2.25$
$a_2=\dfrac{1}{2}\left(a_1+\dfrac{5}{a_1}\right)=\dfrac{1}{2}\left(2.25+\dfrac{5}{2.25}\right)\approx 2.23611111111$
$a_3=\dfrac{1}{2}\left(a_2+\dfrac{5}{a_2}\right)=\dfrac{1}{2}\left(2.23611111111+\dfrac{5}{2.23611111111}\right)\approx 2.23606797792$
$a_4=\dfrac{1}{2}\left(a_3+\dfrac{5}{a_3}\right)=\dfrac{1}{2}\left(2.23606797792+\dfrac{5}{2.23606797792}\right)\approx 2.2360679775$
$a_4=\dfrac{1}{2}\left(a_3+\dfrac{5}{a_3}\right)=\dfrac{1}{2}\left(2.2360679775+\dfrac{5}{2.2360679775}\right)\approx 2.2360679775$
$a_5=\dfrac{1}{2}\left(a_4+\dfrac{5}{a_4}\right)=\dfrac{1}{2}\left(2.2360679775+\dfrac{5}{2.2360679775}\right)\approx 2.2360679775$
So $\sqrt 5\approx 2.236$
