How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?
 A: In binary, this is
$$1+1+10+100+...+1\underbrace{00...00}_{n-2\text{ times}}=1+\underbrace{11...11}_{n-1\text{ times}}=1\underbrace{00...00}_{n-1\text{ times}}$$
which in decimal is $2^{n-1}$.
A: What you suggest basically seems to work - I'm going to change the leading value into $n$ rather than $n-1$ for ease of reading:
First, $2^1 = 1+1$ and $2^2 = 1+1+2 = 1+2^0+2^1$
Then using induction, assuming that the summation holds for lower $n$, we can use your observation $2^n = 2^{n-1} + 2^{n-1}$ and change the first $2^{n-1}$ into the summation form, giving  $2^{n} = 1+1 + 2 + \cdots + 2^{n-2} + 2^{n-1}$ as required.
A: Let me suggest 4 ways:

*

*$ a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ ... + b^{n-1}) \Rightarrow 2^{n} = 2^{n} - 1 + 1 = 1+2+ ... +2^{n-1}  +1 $


*Considering all subsets from set with $ n>1 $ elements and considering map between subsets and binary numbers


*Mathematical Induction


*Newton's binomial theorem
A: It's because it's a geometric progression. Since
$$1+p_i+{p_i}^2+··· +{p_i}^{a_i} =  \dfrac{{p_i}^{a_i+1}-1}{p_i-1}. $$
you just add an extra one at the start of the progression.
A: Let $$x = \sum_{m=2}^{n}2^{m-2},$$ and note that $$1 + x = 1 + 1 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{1}.$$  Then, multiplying $(1)$ by $2$, we have $$2+2x = 2 + 2 + 4 + \dotsb + 2^{n-2} + 2^{n-1}\tag{2}.$$  We can rewrite $(1)$ as $$1 + x = 2 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{3}.$$  Subtracting $(3)$ from $(2)$ gives us $$1 + x = 2^{n-1},$$ which is the same as $$2^{n-1} = 1 + 1 + 2 + 4 + \dotsb + 2^{n-2}.$$
A: Let $S:=1+2+4+\dots+2^{n-2}$.
Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.
