# Show that $|c_n-c_{n+1}|=2^{-n-2}(b_0-a_0)$. [closed]

Let $c_n={1\over 2}(a_n+b_n)$, $r=\lim_{n\to \infty}c_n$, and $e_n=r-c_n$. Here $[a_n,b_n]$, with $n\geq 0$, denotes the successive intervals that arise in the bisection method when it is applied to a continuous function $f$. Show that $|c_n-c_{n+1}|=2^{-n-2}(b_0-a_0)$.

I'm having a tough time showing this. Any hints or solutions are greatly appreciated.

## closed as off-topic by heropup, Daniel Robert-Nicoud, user147263, Daniel W. Farlow, ShaileshApr 22 '16 at 0:11

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Daniel Robert-Nicoud, Community, Daniel W. Farlow, Shailesh
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• All the letters might be confusing you. Try drawing a picture to see what this means. Then, to make it rigorous, consider a proof by induction. – Reveillark Oct 2 '15 at 1:58

$c_n$ is the mid point of $[a_n,b_n]$.
$c_{n+1}$ is the mid point of either $[a_n, c_n]$ or $[c_n,b_n]$.
Compute $|c_n-c_{n+1}|$ in terms of $a_n,b_n$.
$|c_n-c_{n+1}| = {1 \over 4} (b_n-a_n)$.
You know that $b_{n+1}-a_{n+1} = {1 \over 2} (b_n-a_n)$, hence
$b_n-a_n = {1 \over 2^n} (b_0-a_0)$.