Using direct proof, prove that $2^0 + 2^1 + 2^2 + \dots + 2^n = 2^{n+1} - 1$ (no summation, theorems, or induction allowed)

Question

I'm reading Book of Proof for a refresher on mathematical proofs and came across this problem:

If $n \in \mathbb{N}$, then $2^0 + 2^1 + 2^2 + 2^3 + \dots + 2^n = 2^{n+1} - 1$.

I know there have been similar questions on the site asking to prove this problem by induction, but I don't want to use induction because the book has not yet introduced it (even though I'm familiar, but not comfortable, with induction proofs). In fact, my book's answer key uses a direct proof:

Proof. We use direct proof. Suppose $n \in \mathbb{N}$. Let $S$ be the number:

$S = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \dots + 2^n$ (1)

In what follows, we will solve for $S$ and show $S = 2^{n+1} - 1$. Multiplying both sides of (1) by $2$ gives

$2S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + \dots + 2^{n+1}$ (2)

Now subtract Equation (1) from Equation (2) to obtain $2S - S = -2^{0} + 2^{n+1}$, which simplifies to $S = 2^{n+1} - 1$. Combining this with Equation (1) produces $2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \dots + 2^n = 2^{n+1} - 1$, so the proof is complete.

I don't follow how the subtraction was performed. I think I'm very close to understanding this, but there's some key piece missing that's preventing me from articulating this proof.

I understand it a little better if I line up $S$ and $2S$ above each other diagrammatically, in which case it's clear that multiplying $S$ by $2$ is the same as shifting the bits to the right, such that we lose the $2^0$ we had before ($-1$) but gain an extra $2^{n+1}$ that was not there before, such that the total difference ends up being $2^{n+1} - 1$.

$S = \color{red}{2^0} + \color{blue}{2^1 + 2^2 + 2^3 + 2^4 + \dots + 2^{n-1} + 2^n}$

$2S = \color{blue}{2^1 + 2^2 + 2^3 + 2^4 + 2^5 \dots + 2^{n-1} + 2^n} + \color{red}{2^{n+1}}$

How can I express this intuitive understanding more formally? What am I missing? I'm having trouble actually performing the subtraction step.

Equivalently, I know I can translate this problem into one where we're dealing with binary strings, showing that $111...111$ is the same as $1000...000$ (with a leading $1$ in the $n+1$th bit) minus $0000...001$ (one), which produces a binary string with a $1$ in every place from $0$ to $n$. Is that an acceptable proof?

Answer 1

Telescoping

Render

$2^{n+1}=2×2^n=2^n+2^n;\therefore 2^n=2^{n+1}-2^n$

So

$2^0+2^1+2^2+...2^n=(2^1-2^0)+(2^2-2^1)+...+(2^{n+1}-2^n)$

and then cancel the $+2^1$ and $-2^1$ terms, then the $+2^2$ and $-2^2$ terms, and so on through the $+2^n$ and $-2^n$ terms, and you're left with just $+2^{n+1}$ and $-2^0=-1$ making up the sum.

Answer 2

I kind of answered my own question when asking it. Figured I'd formally post an answer with the colored highlights:

$S = \color{red}{2^0} + \color{blue}{2^1 + 2^2 + 2^3 + 2^4 + \dots + 2^{n-1} + 2^n}$

$2S = \color{blue}{2^1 + 2^2 + 2^3 + 2^4 + 2^5 \dots + 2^{n-1} + 2^n} + \color{red}{2^{n+1}}$

(Just distribute the $2$ to every term in $S$, effectively raising each term's power by one. This is identical to taking an $n$-bit binary string and performing a bitwise shift of one place to the left.)

Then, take advantage of the fact that $2S - S = S$:

$2S - S = 2^{n+1} - 2^0 = 2^{n+1} - 1 = S$

Basically, all the blue terms cancel out when performing the subtraction.