Given that the sequence $\left\{a_n\right\}$ satisfies $a_0 \ne 0,1$ and $$a_{n+1}=1-a_n(1-a_n)$$ $$a_1=1-a_0$$
Find the value of $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$$
We actually get the terms of the sequence as:
$$a_1=1-a_0$$
$$a_2=1-a_1a_0$$
$$a_3=1-a_2a_1a_0$$
$$\vdots$$
$$a_{n+1}=1-a_na_{n-1}a_{n-2}\cdots a_0$$
Any way from here?
Best Answer
For $n\ge 1$, we have $a_{n+1}-1=a_n(a_n-1)$. It follows that if $a_n\notin\{0,1\}$, then $a_{n+1}\notin\{0,1\}$, but since $a_0\notin\{0,1\}$, we see that $a_n\notin\{0,1\}$ for all $n$. Now $$\frac1{a_{n+1}-1}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n}$$ for all $n\ge 1$. That is $$\frac1{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$ if $n\ge 1$, so $$\frac1{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=\frac1{a_1-1}-\frac{1}{a_{n+1}-1}$$ Since $a_1=1-a_0$, we get $a_1-1=-a_0$. Thus $\frac{1}{a_1-1}=-\frac1{a_0}$ and $$\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=-\frac{1}{a_{n+1}-1}.$$ From your calculation, $a_{n+1}-1=-a_0a_1a_2\cdots a_n$, so $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}\right)=\frac{a_{n+1}-1}{a_{n+1}-1}=1.$$