Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
$\sum\limits_{i=1}^n 2^{i-1} = \sum\limits_{i=0}^{n-1} 2^{i}= 2^{n}-1$
What's confusing to me about this problem is the fact that it has two equalities rolled into one. Do I prove the two sums, $\sum\limits_{i=1}^n 2^{i-1} = \sum\limits_{i=0}^{n-1} 2^{i}$, equivalent to one another and then prove $2^{n}-1$ equivalent to one of them? Does the order matter? How do you even prove two sums equivalent (using sigma notation)?