Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to ${2^n}{-1}$.

Question

Please explain in quotations!

"Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to $2^n-1$." In a very simple case, for $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.

Answer 1

HINT: Suppose $$\sum_{i=0}^{n-1} 2^i=2^n-1$$ is true, then $$\sum_{i=0}^{n}2^i=\sum_{i=0}^{n-1}2^i+2^{n}=2^{n}-1+2^{n}=2\times2^n-1=2^{n+1}-1$$

Answer 2

They mean to the formula $2^0+2^1+2^2+\ldots+2^{n-1}=2^n-1$ which is a particular case of $$ q^0+q^1+q^2+\ldots+q^{n-1}=\frac{q^n-1}{q-1}\quad(q\neq1) $$ for $q=2$.

Answer 3

Since we don't know what $1+2+4+\cdots+2^{n-1}$ is yet, we'll call that number $S$. So we have $S=1+\cdots+2^{n-1}$. Multiplying by $2$ on both sides, we get $$ 2S=2+4+8+\cdots+2^{n-1}+2^n $$ Here comes the trick: calculate $2S-S$: $$ \begin{align} 2S-S=&2+4+8+\cdots+2^{n-1}+2^n\\-(1+&2+4+8+\cdots+2^{n-1}) \end{align} $$ and we see that all but two of the terms disappear, and we're left with $2S-S=2^n-1$.

Answer 4

If you’re familiar with binary, simply note that

$$2^n-1 = {1\underbrace{000\dots00}_{\text{$n$ zeros}}}\text{$_2$} - 1 = \underbrace{111\dots11}_{\text{$n$ ones}}\text{$_2$}$$

Answer 5

$1 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1}=$

$1 + 1 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1} - 1=$

$2 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1} - 1=$

$4+ 4 + 8 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$

$8 + 8 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$

$16 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$

.......

$2^j + 2^j + 2^{j+1}+ ...... + 2^{n-2} + 2^{n-1} - 1=$

$2^{j+1} + 2^{j+1}+.....+ 2^{n-2} + 2^{n-1} - 1=$

$2^{j+2} + ..... + 2^{n-2} + 2^{n-1} - 1=$

..........

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

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

$2^n - 1$.

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Or another way:

$2^n = 2*2^{n-1} = 2^{n-1} + 2^{n-1}=$

$2^{n-1} + 2*2^{n-2} = 2^{n-1} + 2^{n-2} + 2^{n-2}=$

$2^{n-1} + 2^{n-2} + 2*2^{n-3} = 2^{n-1} + 2^{n-2} + 2^{n-3}+2^{n-3}=$

........

$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 8=$

$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 4=$

$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 2=$

$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 1+1=$

$2^n$

So

$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 1 = 2^n - 1$