Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i  = 2^n - 2
$$
Background
To give you some background I am trying to derive $MaxInt(n) = 2^n-1$ which describes that the maximum integer which can be created using the two's complement where $n$ is the number of bits the integer is encoded by.
How two's complement encodes numbers with 4 bits is explained by the images below:


Therefore:
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^i = (2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2} )
$$
Maybe there is a way of integrating this or simplifying it but I figured that this is a similar problem to sum to N where
$$
\frac{T(n) + T(n)}{2} = T(n) = \sum_{i=1}^{n} n-i+1 = \sum_{i=1}^{n} i 
$$
So following this logic $MaxInt(n)$ is also equal to:
$$
MaxInt(n) = \frac{MaxInt(n) + MaxInt(n)}{2} 
$$
Since
$$
(2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2}) = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Then
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^{n-2-i} = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Putting it all together:
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{n-2-i} + \sum_{i=0}^{n-2} 2^i}{2}
$$
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i}{2}
$$
Which is when I got stuck, cheating with wolfrom alpha I found that
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i  = 2^n - 2
$$
But I don't know why. If you see a better alternative way (i.e. not using sum to N method) of deriving $MaxInt(n) = 2^n-1$ please let me know.
Thanks for reading.
 A: The thing you've asked to show isn't too hard:
\begin{align}
\sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right)  
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \sum_{i=0}^{n-2}\left(2^i\right) \\
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(1 + 2 + \ldots + 2^{n-2}\right) \\
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(2^{n-1} - 1\right) \\
\end{align}
where that last thing comes form the formula for the sum of a geomtric series, which I think you probably know. Now let's simplify the left-hand term...
\begin{align}
\sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right)  
&= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(2^{n-1} - 1\right) \\
&= \sum_{i=0}^{n-2} \left(2^{(n -2)-i)}\right) + \left(2^{n-1} - 1\right) \\
&= \left(2^{n -2} + 2^{n-3} + \ldots + 2^0\right) + \left(2^{n-1} - 1\right) \\
\end{align}
which we recognize as another geometric series, written in reverse order; the sum there gives us
\begin{align}
\sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right)  
&= \left(2^{n -2} + 2^{n-3} + \ldots + 2^0\right) + \left(2^{n-1} - 1\right) \\
&= \left(2^{n -1} - 1\right) + \left(2^{n-1} - 1\right) \\
&= 2 \cdot 2^{n -1} - 2 \\
&= 2^{n} - 2.
\end{align}
Quick proof for the geometric series: 
If we expand
$$
U = (1 - a) (1 + a + a^2 + \ldots a^k)
$$
with the distributive law, and then gather like terms via the commutative law for addition, we get this:
\begin{align}
U 
&= (1 - a) (1 + a + a^2 + \ldots + a^{k-1} + a^k)\\
&= 1 \cdot (1 + a + a^2 + a^{k-1} + \ldots a^k) - a \cdot (1 + a + a^2 + \ldots + a^{k-1} + a^k)\\
&= (1 + a + a^2 + \ldots + a^{k-1} + a^k) - (a + a^2 + a^3 + \ldots + a^k + a^{k+1})\\
&= 1 + (a + a^2 + \ldots a^k) - (a + a^2 + a^3 + \ldots + a^k) - a^{k+1})\\
&= 1  - a^{k+1})
\end{align}
so we have that
$$
1-a^{k+1} = (1-a) (1 + a + \ldots + a^k)
$$
hence (for $a \ne 1$), 
$$
1 + a + \ldots + a^k = \frac{1-a^{k+1}}{1-a},
$$
which is the formula for the sum of a finite geometric series whose ratio is not 1. For an infinite series whose ratio has absolute value less than 1, the infinite sum turns out to be $\frac{1}{1-a}$, by the way, but this requires a careful definition of a sum for an infinite series. 
A: First notice that for any $m$, $\sum_{i=0}^{m-1} 2^i = 2^m-1$. To see this
(and really this is the geometric series proof applied to your situation), let $$S=\sum_{i=0}^{m-1} 2^i=1+2+4+8+\cdots+2^{m-1}.$$ Then \begin{eqnarray*} 2S&=&2+4+8+\cdots+2^{m-1}+2^m\\ &=& (-1+1)+2+4+\cdots+2^{m-1}+2^m\\&=& -1+(1+2+4+\cdots+2^{m-1})+2^m\\ &=& -1 + S + 2^m.\end{eqnarray*}
So $2S=-1+S+2^m$ and so $S=2^m-1$.
Now your sum is $$\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = \sum_{i=0}^{n-2} 2^{(n-2)-i} + \sum_{i=0}^{n-2} 2^i.$$
But $$\sum_{i=0}^{n-2} 2^{(n-2)-i} = \sum_{i=0}^{n-2} 2^i.$$ Here, the left-side is the sum $2^{n-2}+2^{n-3}+\cdots + 2+1$ and the right-side just expresses this sum in the reverse order $1+2+\cdots+2^{n-3}+2^{n-2}$.
So putting this together you have \begin{eqnarray*} \sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i &=& \sum_{i=0}^{n-2} 2^{(n-2)-i} + \sum_{i=0}^{n-2} 2^i\\ &=& \sum_{i=0}^{n-2} 2^{i} + \sum_{i=0}^{n-2} 2^i\\ &=&2\sum_{i=0}^{n-2} 2^i \\ &=& 2(2^{n-1}-1) \\ &=& 2^{n}-2\end{eqnarray*}
A: $$\large\begin{align}
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i  &=\sum_{\begin{matrix}j\ +\ k\ =\ n-2\\ 0\;\le\; j,\ k\;\le\; n-2\end{matrix}}2^j+2^k\\\\
&=2\sum_{r=0}^{n-2}2^r\\
&=\sum_{r=1}^{n-1}2^r\\
&=\sum_{r=1}^{n-1}2^{r+1}-2^r\\
&=2^n-2\qquad\text{by telescoping}\quad \blacksquare
\end{align}$$
A: Applying the proof of the geometric series to this particular case:
$$
MaxInt(n) = S(n) = \sum_{i=0}^{n-2} 2^i = 2^0 + 2^1 ... + 2^{n-3} + 2^{n-2}
$$
Adding up each term in the series:
$$
2S(n) = (2^0 + 2^0) + (2^1 + 2^1) ... + (2^{n-3} + 2^{n-3}) + (2^{n-2} + 2^{n-2})
$$
Simplifying:
$$
2S(n) = 2^1 + 2^2 ... + (2 \cdot \frac{2^{n}}{2^3}) + (2 \cdot \frac{2^{n}}{2^2})
$$
$$
2S(n) = 2^1 + 2^2 ... + \frac{2^{n}}{2^2} + \frac{2^{n}}{2^1}
$$
$$
2S(n) = {\color{Blue} {2^1 + 2^2 ... + 2^{n-2}} } + 2^{n-1}
$$
Notice what is in blue is actually $S(n)$ except it is missing the first term, lets call this part in blue $y$:
$$
2S(n) = {\color{Blue} {y} } + 2^{n-1}
$$
$$
{\color{Blue} {y} } = S(n) - 2^0
$$
Substituting in:
$$
2S(n) = {\color{Blue} {S(n) -2^0} } + 2^{n-1}
$$
$$
S(n) = 2^{n-1} - 2^0
$$
$$
= 2^{n-1} - 1
$$
$$
\therefore MaxInt(n) = \sum_{i=0}^{n-2} 2^i = 2^{n-1} - 1
$$
The above is the equation desired by the background question, but the original question asks for $2S(n)$
$$
2S(n) = 2 \sum_{i=0}^{n-2} 2^i 
$$
$$
=  2^1 \cdot (2^{n-1} -  1 ) 
$$
$$
= 2^{n-1+1} - 2 
$$
$$
= 2^{n} - 2
$$
A: As, reversing the order,
$$\sum_{i=0}^{n-2}2^{i-n+2}=\sum_{i=0}^{n-2}2^i$$
the global sum is
$$2(2^{n-1}-1).$$
