Evaluate $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}$ when you see $15$ fraction lines.
I have solved this problem but using a quater calculating I come from down to up 15 times and found the answer $\frac{987}{610}$. And the time of exam was an hour It should be an easier way to solve it.That takes about 5 minutes.
update1:The last fraction is:$1+\frac{1}{1}$
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Let $f(x) = 1 + \frac 1x$. We denote $f^k(x) = \overbrace{f(\cdots(f}^{k}(x))\cdots)$.
It would seem, then, that you are calculating $f^{15}(1)$. Note that $$ f(a/b) = 1 + \frac ab = \frac{a + b}{b}. $$ Thus, we can calculate $f^{15}(1)$ efficiently by the process $$ \frac 11 \mapsto \frac{2}{1} \mapsto \frac 32 \mapsto \frac 53 \mapsto \cdots \mapsto \frac{987}{610} \mapsto \frac{1597}{987} $$ this process is made even easier if you happen to know the Fibonacci numbers.
After the first two or three steps you might notice that numerator and denominator are consecutive Fibonacci numbers. Let $\frac{a_n}{b_n}$ be the value with $n$ horizontal lines. Then
$$\frac{a_{n+1}}{b_{n+1}}=1+\frac1{a_n/b_n}=1+\frac{b_n}{a_n}=\frac{a_n+b_n}{a_n}\;:$$
the new numerator is the sum of the old numerator and denominator, and the new denominator is the old numerator. Thus,
$$\frac{a_n}{b_n}=\frac{F_{n+1}}{F_n}\;,$$
and you can easily calculate the first $16$ Fibonacci numbers by hand using the recurrence $F_n=F_{n-1}+F_{n-2}$. Alternatively, you can use the fact that $F_n$ is the integer closest to
$$\frac{\varphi^n}{\sqrt5}\;,$$
where $\varphi=\frac12\left(1+\sqrt5\right)$.