Fractional Equations A

This represents a continued fraction of period 2.

$x_1 + \frac{1}{x_2} = 4 \Rightarrow x_1 = 4 - \frac{1}{x_2} \\ x_2 + \frac{1}{x_3} = 1 \Rightarrow x_2 = 1 - \frac{1}{x_3} \\ x_3 + \frac{1}{x_4} = 4 \Rightarrow x_3 = 4 - \frac{1}{x_4} \\ x_4 + \frac{1}{x_5} = 1 \Rightarrow x_4 = 1 - \frac{1}{x_5} \\ x_1 + \frac{1}{1 - \frac{1}{4 - x_1}} = 4 \Rightarrow x_1 = 2$

Substituting $x_1 = 2$ into the first equation gives $x_2 = \frac{1}{2}$ and by periodicity $x_{2k+1} = 2, x_{2k} = \frac{1}{2}$ for $k \epsilon N$

Therefore $x_1 + x_2 + x_3 + x_4 + ... + x_{100} = (2 + \frac{1}{2})50 = \boxed{125}$

By A.M.-G.M. $x_1 \leq 4x_2$ , $x_2 \leq x_3$ , $x_3 \leq 4x_4$ , $\cdots$ , $x_{100} \leq x_1$

$\therefore$ $x_1=x_3=x_5=\cdots=x_{99} = x_1$ and $x_2=x_4=x_6=\cdots=x_{100} = \frac{x_1}{4}$

$x_1 +\frac{4}{x_1} =4 \implies x_1=2$

$\therefore x_1+x_2+\cdots + x_{100} = \frac{250x_1}{4} =125.$