Proof By Induction Help? I've been working through proof by induction and i'm stuck on this question. Can somebody provide some help?
$$ 2^n-1=\sum_{i=0}^{n-1}2^i\text{ for }n\ge 1$$
 A: First, show that this is true for $n=1$:


*

*$2^1-1=\sum\limits_{i=0}^{1-1}2^i$


Second, assume that this is true for $n$:


*

*$2^n-1=\sum\limits_{i=0}^{n-1}2^i$


Third, prove that this is true for $n+1$:


*

*$2^{n+1}-1=2\cdot2^n-1$

*$2\cdot2^n-1=2\cdot2^n-2+1$

*$2\cdot2^n-2+1=2\cdot(2^n-1)+1$

*$2\cdot(2^n-1)+1=1+2\cdot(2^n-1)$

*$1+2\cdot(2^n-1)=1+2\cdot\sum\limits_{i=0}^{n-1}2^i$ assumption used here

*$1+2\cdot\sum\limits_{i=0}^{n-1}2^i=1+\sum\limits_{i=1}^{n}2^i$

*$1+\sum\limits_{i=1}^{n}2^i=2^0+\sum\limits_{i=1}^{n}2^i$

*$2^0+\sum\limits_{i=1}^{n}2^i=\sum\limits_{i=0}^{n}2^i$
A: Hint : $2^{n+1}-1=2(2^n-1)+1$.
A: Guest has given a very good hint. Alternatively, you can generalize: 
For any $x,y \in \Bbb{R}$ then $$x^n-y^n = (x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$ If you prove this by induction, then your proof follows as a special case of $x=2, y=1$. Your original proof is probably easier than this one, but this one is a good result to have regardless.
A: Although the hint by Guest is quite good, as graydad pointed out, I imagine you are probably struggling with some summation manipulation (or am I wrong?). Here's a brief sketch (I imagine you can fill in all of the fine details and make the argument flow like a written proof should):
\begin{align}
2^{k+1}-1 &= 2\cdot(2^k-1)+1\\[1em]
          &= 2\cdot\sum_{i=0}^{k-1}2^i + 1\\[1em]
          &= \sum_{i=1}^k2^i + 1\\[1em]
          &= \sum_{i=0}^k2^i.
\end{align}
Does that make more sense now?
