Sum of Powers of 2/Proof 1

Theorem

Let $n \in \N_{>0}$ be a (strictly positive) natural number.

Then:

\(\text {(1)}: \quad\)

\(\ds 2^n - 1\)

\(=\)

\(\ds \sum_{j \mathop = 0}^{n - 1} 2^j\)

\(\ds \)

\(=\)

\(\ds 1 + 2 + 2^2 + 2^3 + \dotsb + 2^{n - 1}\)

Proof

From Sum of Geometric Sequence:

$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$

The result follows by setting $x = 2$.

$\blacksquare$