Why is: $1+2+2^2+\cdots+2^{n-2} = 2^{n-1}-1$?

Question

I know that this might be elementary, but I don't know how to prove this equality, some hint? I tried to extract the common two:

$$ 1 + 2^1+2^2 + \cdots + 2^{n-2} = 1 + 2(1+2+\cdots+2^{n-3}) $$

But this didn't help

Answer 1

The sum $1+2+2^2+\cdots+2^{n-2} $ doesn't change if we multiply it by $1$ i.e. $2-1$:

\begin{eqnarray}1+2+2^2+\cdots+2^{n-2} &=& (1+2+2^2+\cdots+2^{n-2})(2-1)\\ &=& (\color{red}{2+2^2+2^3+\cdots+2^{n-2}}+2^{n-1})-(1+\color{red}{2+2^2+2^3+\cdots+2^{n-2}})\\ &=&2^{n-1}-1 \end{eqnarray}

Answer 2

Note that $$ \sum_{k=0}^{n-2}2^k=\sum_{k=0}^{n-2}(2^{k+1}-2^k)=2^{n-1}-1 $$ since the last series is telescoping,

Answer 3

For a proof without words, draw a picture of the interval from $0$ to, say, $64$. Divide it in half to make two subintervals of length $32$, divide one of those in half, and so on. Look at what's left when you divide the last subinterval of length $2$.

I would accept this as a convincing argument (others might not) unless the assignment explicitly asked for a proof by induction.

Answer 4

As suggested by Fakemistake in the comments, the sum of a geometric series is $$\sum_{k=0}^{m-\color{red}1}a^k=\dfrac{a^m-1}{a-1}.$$

Apply this with $a=2$ and $m=n-1$.