1. ## Fill in missing step please!

Hi, I am probably an idiot for missing this but im sick of staring at this problem i thought i would come here to be enlightened. anyways
phase 1
assuming n,m > 0 and real numbers and m > n
$\frac{1}{2^n}\left(1+\frac{1}{2}+...+\frac{1}{2^{m-n-1}}\right)$
phase 2
$\frac{1}{2^n}\cdot\frac{\left(1-(\frac{1}{2})^{m-n-1+1}\right)}{1-\frac{1}{2}}$

there are all equal statements im told, but I have no clue the 'trick' to widdle this down from phase 1 to phase 2. thanks in advance

2. Let $S = 1 + r + r^2 + \cdots + r^n$. Then $1+rS = 1 + r + r^2 + \cdots + r^n + r^{n+1} = S + r^{n+1}$ and rearranging gives $S = {1-r^{n+1}\over1-r}$. Use this (with r=1/2) to get from phase 1 to phase 2.

3. Originally Posted by fizzle45
Hi, I am probably an idiot for missing this but im sick of staring at this problem i thought i would come here to be enlightened. anyways
phase 1
assuming n,m > 0 and real numbers and m > n
$\frac{1}{2^n}\left(1+\frac{1}{2}+...+\frac{1}{2^{m-n-1}}\right)$
phase 2
$\frac{1}{2^n}\cdot\frac{\left(1-(\frac{1}{2})^{m-n-1+1}\right)}{1-\frac{1}{2}}$

there are all equal statements im told, but I have no clue the 'trick' to widdle this down from phase 1 to phase 2. thanks in advance
If n,m are, in addition, integers, then this is just an application of the formula for the sum of a geometric series:
$1+q+q^2+\cdots+q^N=\frac{1-q^N}{1-q}$
were $q=\frac{1}{2}$ and $N=m-n-1$

4. thank for quick response guys very helpful!