The sequence $\{a_n\}$ is defined by formula $a_0=4$ and $a_{n+1}=a_n^2-2a_n+2$ for $n\ge0$. Let the sequence $\{b_n\}$ is defined by formula $b_0=\frac12$ and $b_n=\frac{2a_0a_1a_2…a_{n-1}}{a_n}\forall n\ge1$. Write the recurrence formula for the sequence $\{b_n\}$ i.e. write $b_{n+1}$ in terms of $b_n$
My Attempt:
$a_{n+1}=(a_n-1)^2+1$
So, $a_1=10, a_2=82, a_3=6562$
$b_1=\frac45, b_2=\frac{40}{41}, b_3=\frac{3280}{3281}$
How to write recurrence formula?
Best Answer
Since $a_{n+1}-1=(a_n-1)^2$ for all $n\ge 0$, it follows that $a_n-1=(a_0-1)^{2^n}=3^{2^n}$, i.e. $a_n=3^{2^n}+1$. Thus, $$ 2a_0a_1\dots a_{n-1}=\prod_{j=0}^{n-1}\left(3^{2^j}+1\right)=(3-1)(3+1)(3^2+1)(3^{2^2}+1)\dots(3^{2^{n-1}}+1)=3^{2^n}-1, $$ so $$ b_n=\frac{3^{2^n}-1}{3^{2^n}+1}, $$ and therefore, $$ \frac{1+b_n}{1-b_n}=3^{2^n}=\left(3^{2^{n-1}}\right)^2=\left(\frac{1+b_{n-1}}{1-b_{n-1}}\right)^2. $$ Thus, $$ b_n=\frac{\left(\frac{1+b_{n-1}}{1-b_{n-1}}\right)^2-1}{\left(\frac{1+b_{n-1}}{1-b_{n-1}}\right)^2+1}=\frac{(1+b_{n-1})^2-(1-b_{n-1})^2}{(1+b_{n-1})^2+(1-b_{n-1})^2}=\frac{4b_{n-1}}{2+2b_{n-1}^2}=\frac{2b_{n-1}}{1+b_{n-1}^2}. $$