If $a_{1}=\frac{1}{1-x}, a_{2}=\frac{1}{1-a_{1}}$, and $a_{n}=\frac{1}{1-a_{n-1}}$, for $n \geq 2, x \neq 1$ and $x \neq 0$, then $a_{107}$ is
(A) $\frac{1}{1-x}$
(B) $x$
(C) $-x$
(D) $\frac{x-1}{x}$
(E) $\frac{1}{x}$

## Part C: Each question is worth 8 credits.
If $a_{1}=\frac{1}{1-x}, a_{2}=\frac{1}{1-a_{1}}$, and $a_{n}=\frac{1}{1-a_{n-1}}$, for $n \geq 2, x \neq 1$ and $x \neq 0$, then $a_{107}$ is
(A) $\frac{1}{1-x}$
(B) $x$
(C) $-x$
(D) $\frac{x-1}{x}$
(E) $\frac{1}{x}$

## Solution

$a_{1}=\frac{1}{1-x}$

$a_{2}=\frac{1}{1-\frac{1}{1-x}}=\frac{(1-x)(1)}{(1-x)\left(1-\frac{1}{1-x}\right)}=\frac{1-x}{1-x-1}=\frac{1-x}{-x}=\frac{x-1}{x}$

$a_{3}=\frac{1}{1-\frac{x-1}{x}}=\frac{x(1)}{x\left(1-\frac{x-1}{x}\right)}=\frac{x}{x-(x-1)}=x$

$a_{4}=\frac{1}{1-x}$

Since $a_{1}=a_{4}$, we conclude $a_{1}=a_{4}=a_{7}=\ldots=a_{3 n-2}=a_{106}$.

Also, $a_{2}=a_{5}=a_{8}=\ldots=a_{3 n-1}=a_{107}$ for $n=36$.

Since $a_{2}=\frac{x-1}{x}$ then $a_{107}=\frac{x-1}{x}$.

ANSWER: (D)

## Part C