Taken from the introductory chapter of "How to Prove it A Structured Approach":
$(2^b - 1)(1 + 2^b + 2^{2b} + ... + 2^{(a - 1)b}$
$=2^b(1 + 2^b + 2^{2b} + ... + 2^{(a - 1)b}) - (1 + 2^b + 2^{2b} + ... + 2^{a - 1)b})$
$=(2^b + 2^{2b} + 2^{3b} + ... + 2^{ab}) - (1 + 2^b + 2^{2b} + ... + 2^{(a - 1)b})$
$=2^{ab} - 1$
A line 3 you take out all $2^b, 2^{2b}, 2^{3b}$ from the right side out of the left side and the only ones that remains are the $2^{ab}$ on the left side, the $-1$ on the right side and the $-2^{(a - 1)b}$ on the right side, so the final result should be:
$2^{ab} - 2^{(a - 1)b} - 1$
Shouldn't it?Yet the author removes the $2^{(a - 1)b}$ at the end of the right side, but how does he cancel it out?
To see the result without ambiguity:
$$\left(2^b - 1\right)\left(1 + 2^b + 2^{2b} + ... + 2^{(a - 1)b}\right)=\left(2^b - 1\right)\sum_{k=0}^{a-1}2^{bk}=\sum_{k=0}^{a-1}2^{b(k+1)}-\sum_{k=0}^{a-1}2^{bk}\\\underbrace{=}_{\text{change of index}} \sum_{k=1}^{a}2^{bk}-\sum_{k=0}^{a-1}2^{bk}=2^{ab}-1$$