An algebra problem by Rahil Sehgal

$\begin{aligned} f_0(x) & = \frac 1{1-x} \\ f_1(x) & = f_0(f_0(x)) = \frac 1{1-\frac 1{1-x}} = \frac {1-x}{-x} = \frac {x-1}{x} = 1-\frac 1x \\ f_2(x) & = \frac 1{1-1+\frac 1x} = x \\ f_3(x) & = f_0(x) \\ \implies f_n (x) & = f_{n-3} (x) = \begin{cases} \dfrac 1{1-x} & \text{if } n \text{ mod 3} = 0 \\ 1 -\dfrac 1x & \text{if } n \text{ mod 3} = 1 \\ x & \text{if } n \text{ mod 3} = 2 \end{cases} \end{aligned}$

$\begin{aligned} \implies f_{100}(3) + f_1\left(\frac 23\right) + f_2\left(\frac 32\right) & = f_1(3) + f_1\left(\frac 23\right) + f_2\left(\frac 32\right) \\ & = 1 - \frac 13 + 1 - \frac 32 + \frac 32 \\ & = \frac 53 \approx \boxed{1.66} \end{aligned}$