How do I solve this recurrence relation?

Question

Given a recursive relation $$a_n = \begin{cases} (1 - 2b_n)a_{n-1} + b_n, & n > 1 \\ \frac{1}{2}, & n =1 \end{cases} $$, how can I expression $a_n$ in term of $b_i, i \in \{1, 2, \dots n\}$?

Answer 1

Suppose that $a_n=\frac{1}{2}$ for some $n$. Then according to the recurrence:

\begin{align*} a_{n+1}&=(1-2b_{n+1})a_n+b_{n+1} \\ &= \frac{1}{2}-b_{n+1}+b_{n+1} \\ &= \frac{1}{2} \end{align*}

Since $a_1=\frac{1}{2}$, by induction this shows that $a_n=\frac{1}{2}$ for all $n$.

Answer 2

Consider the case of $n=2$ with $a_1=\frac{1}{2}$. $$a_2=(1 - 2b_2)a_1 + b_2=(1 - 2b_2)\frac{1}{2}+b_2=\frac{1}{2}$$ You can repeat that for ever and, whatever $b_n$ could be, all $a_n=\frac{1}{2}$.