Bisection Method: How to find upper bound of interval width at n steps in terms of initial interval

Question

In my book on numerical algebra, in the chapter discussing the bisection method for root-finding, I've encountered a practice problem which does not seem very clear to me:

Suppose that $\left|a_n-b_n\right| \le \lambda_n\left|a_{n-1}-b_{n-1}\right|$ for all $n$ with $\lambda_n < 1$. Find an upper bound on $\left|a_n-b_n\right|$ in terms of $\left|a_0-b_0\right|$ and $\lambda = max_{1\le i\le n}\{\lambda_i\}$.

Perhaps it's the terseness of its phrasing, or the lack of any similar examples anywhere in the text, but I'm not quite sure where to begin solving this. I expect the solution should be similar to the formula for the width of the interval at n-steps, $b_n - a_n = 2^{-n}(b_0 - a_0)$, but the lambda throws me off. Could anyone clarify this a bit?

Answer 1

$\left|a_n-b_n\right| \le \lambda_n\left|a_{n-1}-b_{n-1}\right|\le\lambda\left|a_{n-1}-b_{n-1}\right|$

So by applying it $n$ times:

$\left|a_n-b_n\right| \le \lambda^n\left|a_0-b_0\right|$