Solution:
So, we have \(2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}} \), which at first looks rather scary. But we notice a pattern! Because the nested fractions descend forever, the fraction essentially contains itself! So let \(a\) equal this fraction:
\( a = 2+\frac{4}{1+\frac{4}{2+\frac{4}{1+\cdots}}} \)
Thus,
\( a = 2+\frac{4}{1+\frac{4}{a}} \)
So now, we simply need to solve for a.
\(a = 2+\frac{4}{1+\frac{4}{a}} = 2 + \frac{4}{\frac{4+a}{a}}= 2+\frac{4a}{4+a}=\frac{8+2a+4a}{4+a}=\frac{8+6a}{4+a}\)
Getting rid of the fraction by multiplying y \(4+a\), we get
\(4a + a^2 = 8 + 6a \Longrightarrow a^2 - 2a - 8 = 0\)
So now we factor...
\(a^2 - 4a + 2a - 8 = 0\)
\(a(a-4)+2(a-4) = 0\)
\((a+2)(a-4) = 0\)
So \(a\) can be \(-2\) or \(4\). Since we know, by looking at the fraction, that \(a\) must be positive, \(a\), and the value of the fraction, must be \(\boxed{4}\).
Tell me if you have any questions,
