Prove mathematical induction $$\sum_{i=0}^{k-1} 2^i = 2^k-1$$
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A proof can be found towards the end of this answer. – Dilip Sarwate Dec 02 '15 at 04:02
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Assuming you mean $2^k-1$ then
base case $k=1$ then
$$\sum_{i=0}^{1-1}2^i = 1 = 2^1 -1$$
Now $$\begin{align}\sum_{i=0}^{k}2^i & = \sum_{i=0}^{k-1}2^i +2^k \\[0.5ex] & = 2^k -1 + 2^k & \textsf{assuming }\sum_{i=1}^{k-1} = 2^k-1 \\[0.5ex] & = 2^12^k-1 \\[0.5ex] & =2^{k+1}-1 \end{align}$$

Graham Kemp
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