CWMI 2008 P1 wrote: A sequence of real numbers $ \{a_{n}\}$ is defineds by $ a_{0}\neq 0,1$, $ a_1=1-a_0$,$ a_{n+1}=1-a_n(1-a_n)$, $ n=1,2,...$. Prove that for any positive integer $ n$, we have $$ a_{0}a_{1}...a_{n}(\frac{1}{a_0}+\frac{1}{a_1}+...+\frac{1}{a_n})=1$$ Solution: Suppose for some $n \in \mathbb{N}$, we have: $a_{n}=1-\prod_{k=0}^{n-1} a_k$ $\implies$ $a_{n+1}$ $=$ $1-a_n(1-a_n)$ $=$ $1-\prod_{k=0}^n a_k$ $\implies$ $$a_{n+1}=1- \prod_{k=0}^{n} a_k$$Now when, $n=2$ $\implies$ $\text{ LHS }$ $=$ $a_0a_1+a_1a_2+a_0a_2=a_0a_1+a_2(a_0+a_1)=a_0a_1+1-a_0a_1=1$. Suppose for some $n \in \mathbb{N}$, it holds that: $$\sum_{k=0}^n \left( \frac{\prod_{i=0}^n a_i }{a_k} \right) =a_1a_2 \ldots a_n + a_0a_2a_3 \ldots a_n + a_0a_1a_3 \ldots a_n +\ldots + a_0a_1a_2\ldots a_{n-1}=1$$Then, $$\sum_{k=0}^{n+1} \left( \frac{\prod_{i=0}^{n+1} a_i }{a_k} \right) =\left( \prod_{k=0}^{n} a_k \right) +a_{n+1} \left( \sum_{k=0}^n \left( \frac{\prod_{i=0}^n a_i }{a_k} \right) \right) = a_{n+1} +\prod_{k=0}^{n} a_k =1 \qquad \blacksquare$$