Largely wrong

Calculating a couple more terms, we find that $A_3=1$ $A_4=-1$ Now, we use induction to prove that $A_{2n}=-1$ for $n\ge 2$ . The base case $n=2$ is true.

Assume the result is true for $n=k$ , such that $A_{2k}=-1$ .

Let $A_{2k-1}A_{2k-2}\cdots A_1=P$ .

Then, $A_{2k+1}=A_{2k}-P=-1-P$ Now, $A_{2k+2}=A_{2k+1}-A_{2k}P=(-1-P)-(-P)=-1$ Therefore, the result is true for $n=k+1$ , and by induction, for all $n\ge 2$ .

Since $1000$ is even, $|A_{1000}|=\boxed{1}$ .

We can list out a few terms: $A_1 = 1$ $A_2 = 2$ $A_3 = 1$ $A_4 = -1$ $A_5 = -3$ $A_6 = -1$ $A_7 = -7$ $A_8 = -1$ $A_9 = -43$ $A_{10} = -1$ $A_{11} = -1807$

Note that starting from $A_4$ , every other term is $-1$ . Let us prove that this holds in the general case; that is, for all integers $n$ such that $n > 1$ , $A_{2n} = -1$ . We will do so by induction.

The base case is $n = 2$ . We have already shown that $A_4 = 1$ , which proves this case.

Now for the inductive step. We assume that for an integer $k$ , $A_{2k} = -1$ . We need to show that $A_{2(k+1)} = A_{2k+2} = -1$ .

First, note that $A_{2k+1} = A_{2k} - A_1A_2\cdots A_{2k-1} = -1 - A_1A_2\cdots A_{2k-1}.$ Then we see that $A_{2k+2} = A_{2k+1} - A_1A_2\cdots A_{2k-1}A_{2k} = -1 - A_1A_2\cdots A_{2k-1} - (-A_1A_2\cdots A_{2k-1}) = -1,$ so our induction is complete.

Thus, we have shown that for all integers $n > 1$ , $A_{2n} = -1$ . When we plug in $n = 500$ , we obtain $A_{1000} = -1$ ; so, the answer is $|A_{1000}| = \boxed{1}$ .

By listing the first few terms of the sequence, we see that: $A_3 = 1$ $A_4 = -1$ $A_5 = -3$ $A_6 = -1$ $A_7 = -7$ $A_8 = -1$ Notice that: $\forall n \text{ such that } n \equiv 0\pmod{2}, \quad A_n = -1$ Therefore, we can see that $A_{1000} = -1$ so: $|A_{1000}| = \fbox{1}$

We will claim that $x_n = -((-\omega)^{n+1}+(-\omega^2)^{n+1})$

where $\omega$ is the 3rd root of unity.

We can proove our claim as follows-

setting $n=3$ we can define the characteristic polynomial of the recurrence as follows -

$x^3=x^2-x \Rightarrow x^2 -x+1=0 \Rightarrow x = -\omega , -\omega^2$

Hence the recurrence will be -

$x_n = c_1(-\omega)^n + c_2(-\omega^2)^n$

Now setting $x_1=1$ and $x_2=2$ , we get $c_1=-\omega$ and $c_2=-\omega^2$

Hence, $x_n = -((-\omega)^{n+1}+(-\omega^2)^{n+1})$

Now $x_{1000} = -(-\omega^{1001} - \omega^{2002})$

Since we know that $\omega^3 = 1$ and $1+ \omega + \omega^2 = 0$

we get, $x_{1000} = -(-\omega^2 - \omega) = -1$ So, $|x_{1000}| = 1$

Let's look at the first few terms of the sequence: $$A 1=1$$ $$A 2=2$$ $$A 3=1$$ $$A 4=-1$$ $$A 5=-3$$ $$A 6=-1$$ $$A 7=-7$$ $$A 8=-1$$ It seems that if $n$ is even and positive, then $A_n$ is -1. This is not terribly hard to prove, therefore: $\boxed{\mid A_{1000}\mid =11}$

Computing the first few terms, we find that $A_{3} = 1, A_{4} = -1, A_5 = -3$ and $A_6 = -1$ . Note that since $A_4 = A_6$ , $A_{2n} = A_4 = -1$ for all $n$ greater than $3$ . Since $1000 \equiv 0 \pmod {2}$ , $A_{1000} = A_4 = -1$ , so the absolute value of $A_{1000}$ is $\boxed{001}$ .

A3 = 1 A4 = -1 A5 = -3 A6 = -1 A7 = -7 A8 = -1 A9 = -43 A10 = -1 . . . . A1000 = -1 because every Ai , i = even number, the value of this Ai = -1 so l A1000 l = 1 .