Sequence Buster

Note that $a_n \; = \; \frac{1-a_{n+1}}{1-a_n} \hspace{2cm} n \ge 1$ so that $\prod_{n=0}^N a_n \; = \; a_0 \prod_{n=1}^N \frac{1-a_{n+1}}{1-a_n} \; = \; a_0 \frac{1 - a_{N+1}}{1-a_1} \; = \; 1 - a_{N+1}$ for all $N \ge 0$ . Also $\frac{1}{1-a_{n+1}} - \frac{1}{1-a_n} \; = \; \frac{1}{a_n(1-a_n)} - \frac{1}{1-a_n} \; = \; \frac{1}{a_n} \hspace{2cm} n \ge 1$ so that $\sum_{n=0}^N \frac{1}{a_n} \; = \; \frac{1}{a_0} + \sum_{n=1}^N\left(\frac{1}{1-a_{n+1}} - \frac{1}{1-a_n}\right) \; = \; \frac{1}{a_0} + \frac{1}{1-a_{N+1}} - \frac{1}{1-a_1} \; = \; \frac{1}{1 - a_{N+1}}$ for all $N \ge 0$ , and hence $\prod_{n=0}^Na_n \times \sum_{n=0}^N \frac{1}{a_n} \; = \; \boxed{1}$ for all $N \ge 0$ .

First inductively we can show, $a_{n+1}=1-a_0.a_1...a_n$ Then we can inductively show, $\sum_0^n \frac{1}{a_i} =\frac{1}{a_0...a_n}$ Hence ,the answer is 1.

From the first few $a_n$ , we note that $a_n = 1 - p_{n-1}$ , where $p_n = \prod_{k=0}^n a_k$ . We can prove this claim by induction . For $n=1$ , $a_1 = 1-a_0 = 1-p_0$ , the claim is true for $n=1$ . Assuming the claim is true for $n$ , then

$a_{n+1} = 1 - a_n(1-a_n) = 1-a_n(1-1+p_{n-1}) = 1-p_n$

The claim is also true for $n+1$ and hence true for all $n \ge 1$ .

We can also prove by induction that $\displaystyle s_n = \sum_{k=0}^n \frac 1{a_k} = \frac 1{p_n}$ . For $n=1$ , $s_1 = \dfrac 1{a_0} + \dfrac 1{a_1} = \dfrac {a_1+a_0}{a_0a_1} = \dfrac {1-a_0+a_0}{p_1} = \dfrac 1{p_1}$ . The claim is true for $n=1$ . Assuming the claim is true for $n$ , then

$s_{n+1} = s_n + \frac 1{a_{n+1}} = \frac {a_{n+1} + p_n}{p_{n+1}} = \frac {1-p_n+p_n}{p_{n+1}} = \frac 1{p_{n+1}}$

The claim is also true for $n+1$ and hence true for all $n \ge 1$ .

Therefore,

$a_0a_1a_2\cdots a_n \left(\frac 1{a_0} + \frac 1{a_1} + \frac 1{a_2} + \cdots + \frac 1{a_n} \right) = p_n \left(\frac 1{p_n} \right) = \boxed 1$

The solution provided by Mark Hennings is great, but I did something that I think I shouldn't have. So I'll share that here.

First, We write $a_{2} = 1-a_{1}+{a_{1}}^{2} = 1-(1-a_{0})+{(1-a_{0})}^{2} = 1-a_{0}+{a_{0}}^2$

Thus, $a_{0}$ and $a_{1}$ are cube roots of $-1$ . That is $a_{0} = -\omega$ and $a_{1} = -{\omega}^{2}$ , You can interchange them but it won't matter because in both cases $a_{2} = 0$ .

Thus all other $a_{i} = 1$ .

Since we need to find the sum of the products of these numbers taking all but one at a time, we can easily see that there is only one term that would be free of $a_{2}$ , and that would be $-\omega \times -{\omega}^{2} \times 1 \times ... \times 1 = \fbox{1}$