Why is it intuitively "obvious" that: $2^0+2^1+2^2+...+2^n=2^{n+1}-1$

Question

Why is it "obvious" that: $$2^0+2^1+2^2+...+2^n=2^{n+1}-1$$

And how about: $$2^{-1}+2^{-2}+2^{-3}+...+2^{-n}=1-2^{-n}$$

I am looking for some intuition about these facts.

Is there a similar pattern when the base is $3$?

Answer 1

$2^{n+1} = 2^n + 2^n = 2^n + 2^{n-1}+ 2^{n-1}$, etc, etc.

Answer 2

In binary we have $2^{k} = 1\underbrace{00\cdots 0}_{k \text{ times}}\ _2$.

When we subtract $1$, we have to borrow, and continue borrowing until the units digit.

$$2^k - 1 = 100\cdots 0_2 - 1_2 = \underbrace{11\cdots 1}_{k \text{ times}}\ _2$$

and

$$\underbrace{11\cdots 1}_{k \text{ times}}\ _2 = 1_2 + 10_2 + 100_2 +\cdots + 1\underbrace{00\cdots 0}_{k-1 \text{ times}} = \sum_{i=0}^{k-1} 2^i$$

giving

$$2^k-1 = \sum_{i=0}^{k-1} 2^i$$

Answer 3

This is like when your car's odometer shows $9999$ (I omit leading zeros): when you do another kilometer (or mile), the rightmost $9$ becomes $0$ and so on for all digits, ending up in $10000$.

Why is that? Because $9$ is the highest digit.

Now think binary, where the highest digit is $\texttt{1}$: $\texttt{0}$, $\texttt{1}$, $\texttt{10}$, $\texttt{11}$, $\texttt{100}$, $\texttt{101}$, $\texttt{110}$, $\texttt{111}$, $\texttt{1000}$, and so on. Look at the numbers with “all ones”: when you add one it's like for the odometer and all digits become $\texttt{0}$, with a $\texttt{1}$ appearing at the far left. Now recall the meaning of a binary number: $$ \texttt{1111}=2^0+2^1+2^2+2^3 $$ and, in general, $$ \underbrace{\texttt{111}\dots\texttt{1}}_{n\text{ digits one}}= 2^0+2^1+\dots+2^{n-1} $$ When you add one to the last number, you get $$ \texttt{1}\underbrace{\texttt{000}\dots\texttt{0}}_{n\text{ digits zero}} =2^n $$ Therefore $2^n=2^0+2^1+\dots+2^{n-1}+1$

Answer 4

Adding another answer to the very good proofs below : you are already familiar with this concept each time you use numbers.

$999_{10} + 1 = 1000_{10}$ is the same than $(9 \times 10^2)+(9 \times 10^1)+(9 \times 10^0) +1 = 10^3$.

If you set $b=9_{10}$ you wrote $(b-1) \times b^2+ (b-1) \times b^1+(b-1) \times b^0 +1 = b^3 = (b-1) ( b^2+b^1+b^0 )$

In base 2, it is the same :

$111_2 = (1 \times 2^2)+(1 \times 2^1)+(1 \times 2^0) +1 = 2^3$.

If you set $b=2_{10}$ you wrote $(b-1) \times b^2+ (b-1) \times b^1+(b-1) \times b^0 +1 = b^3 = b^2+b^1+b^0+1$

Not a proof, just a recall of a knowledge already acquired ...

Answer 5

For a geometric series: $$ q^{n+1}= q^{n}(q-1)+q^n = q^n (q-1)+ q^{n-1}(q-1) + q^{n-1} = ... = (q^n+q^{n-1}+ ...+ 1)(q-1)+1$$

Answer 6

The "base 10" version of this fact certainly seems obvious: \begin{equation} 9999 = 10000 -1. \end{equation}

If you know how to count in base $2$, then your equation is equally obvious. Expressed in binary notation, your equation simply states that \begin{equation} 1111 = 10000 - 1. \end{equation}