Why is $2^{n}-1=\sum\limits_{k=0}^{n-1}2^k$? Recently I discovered that $2^{n}-1=\sum\limits_{k=0}^{n-1} 2^k$. As I don't have a math background, please tell me what this is called and a proof of why this is the case.
 A: Has anyone ever seen this proof anywhere? ( Different to geometric series proof)
in base 2: $$\underbrace{11\cdots 11}_{n\text{}} = \sum_{k=0}^{n-1} 2^k$$ 
the next number after $\underbrace{11\cdots 11}_{n\text{}} $ is ${1\underbrace{0\cdots 0}_{n\text{}}} = 2^n$
so 
$\underbrace{11\cdots 11}_{n\text{}} +1  = 2^n$
or
$\underbrace{11\cdots 11}_{n\text{}} = 2^n -1$
$$\sum_{k=0}^{n-1} 2^k  = 2^n -1$$ 
so in base $x$ we get 
$$\sum_{k=0}^{n-1} x^k  = x^n -1$$ 
with the additional benefit of having defined the next number in any base (integer, complex, matrix etc. )
A: As Hayden has pointed out, the correct expression is $2^n-1=\sum_{k=0}^{n-1} 2^k.$
So to see this, let’s first name the sum $S:=1+2^1+\dots+2^{n-2}+2^{n-1}.$
The ‘trick’ is to multiply the sum by $1$ in a clever way such that we can cancel some terms. Since $2=2-1$, we can say this:
$$
S=1+2^1+\dots+2^{n-2}+2^{n-1}\\
\implies\\
1\times S = \left(2-1\right)\times \left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
S=2\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
-1\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= \left(2^1+2^2+\dots+2^{n-1}+2^{n}\right)\\
-\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= 2^n-1.
$$
Hope this helps!
Stay safe
A: For all positive integer $n$ :$$\begin{align}2^n&=\sum_{k=0}^{n}2^k-\sum_{k=0}^{n-1}2^k&&=(1+2+\cdots+2^{n-1}+2^n)-(1+2+\cdots+2^{n-1})\\[1ex]&=1+\sum_{k=1}^{n}2^k-\sum_{k=0}^{n-1}2^k&&=1+(2+\cdots+2^{n-1}+2^n)-(1+2+\cdots+2^{n-1})\\[1ex]&=1+2\sum_{k=1}^n2^{k-1}-\sum_{k=0}^{n-1}2^k&&=1+2(1+2+\cdots+2^{n-1})-(1+2+\cdots+2^{n-1})\\[1ex]&=1+2\sum_{k=0}^{n-1}2^k-\sum_{k=0}^{n-1}2^k\\[1ex]&=1+\sum_{k=0}^{n-1}2^k&&=1+(1+2+\cdots+2^{n-1})\\[3ex]\therefore\quad\sum_{k=0}^{n-1}2^k&=2^n-1\end{align}$$
A: This is an observation I made myself as a kid and I proved a version of the formula. It's quite easy to do. This is called a geometric series. You start with a first term, then the rule is that the next term is obtained by multiplying the previous term by a fixed constant. The individual terms are called a geometric "sequence", but the sum of all the terms from first to last (in a finite list of terms) is called a geometric series.
The formula is easy to prove. The usual notation for the constant number you multiply each term by is $r$, denoting the "common ratio".
Let's say we have the series: $S(n) = 1 + r + r^2 + r^3 + ...+ r^{n-2} + r^{n-1}$. (series 1)
Note that the argument on the left is $n$ because you're summing $n$ terms on the right.
Now, let's multiply every individual term in the series by $r$. Note that this has the effect of multiplying the entire series by $r$.
$rS(n) = r + r^2 + r^3 + ... r^{n-1} + r^{n}$. (series $2$)
What happens if you take the old series and subtract it from the new one (series $2$ - series $1$)? Note that only the first term in the series $1$ and the last term in series $2$ are "unmatched" and will therefore "survive". All the middle terms vanish.
So you're left with: $rS(n) - S(n)= r^n - 1$
Rearrange that to: $S(n) = \frac{r^n-1}{r-1}$, and you've got your geometric series sum formula. If you put $r = 2$ into that formula, you will reproduce your original observation (with the small correction that others have mentioned), but of course, this is applicable to all $r$.
Note that in the usual statement of the formula, they assume the first term to be $a$ rather than $1$ as I stipulated above. This simply has the effect of mutiplying the entire series by $a$, so the formula is most often stated in texts as $S(n) = \frac{a(r^n-1)}{r-1}$. Not that relevant to your specific question here, but I thought you should know, in order to avoid confusion in the future.
A: Well, here is a simple proof using telescopic sums :
Let $ n $ be a positive integer, we have the following : $$ \sum_{k=0}^{n-1}{2^{k}}=\sum_{k=0}^{n-1}{2^{k}\left(2-1\right)}=\sum_{k=0}^{n-1}{\left(2^{k+1}-2^{k}\right)}=2^{n}-1 $$
A: This is called a geometric series.  One way to prove the formula is mathematical induction.
First we check the formula (corrected the left side with $-1)$  for $n=1$:  indeed, $2^1-1=2^0.$
Now we assume it's true for $n=m$ (i.e., $2^m-1=\sum\limits_{k=0}^{m-1}2^k$), 
and we show it's then true for $n=m+1$:  
$2^{m+1}-1=2^m+2^m-1=2^m+\sum\limits_{k=0}^{m-1}2^k=\sum\limits_{k=0}^{m}2^k.$
Therefore, it's true for $n=1, n=1+1=2, n=2+1=3, n=3+1=4, \dots$
