First show the statement holds for the base case (presumably \(n=1\)):
\(\displaystyle\sum_{i=1}^1 2^{i-1}=2^{1-1}=2^0=1\)
Meanwhile, the right hand side evaluates to \(2^1-1=2-1=1\), so the base case holds.
Now assume the formula holds for \(n=k\); that is,
\(\displaystyle\sum_{i=1}^k 2^{i-1}=2^0+2^1+\cdots+2^{k-1}=2^k-1\)
and use this hypothesis to show that the formula holds for \(n=k+1\). You have
\(\displaystyle\sum_{i=1}^{k+1}2^{i-1}=2^k+\sum_{i=1}^k 2^{i-1}=2^k+2^k-1=2\times2^k-1=2^{k+1}-1\)
so the formula holds and the proof is complete.
