Sum of a finite geometric progression

Statement

Let $x$ be a real number with $x \neq 1$, and let $n$ be a natural number. Then $$\sum_{k=0}^{n-1} x^k = 1 + x + x^2 + \dots + x^{n-1} = \frac{x^n - 1}{x - 1}.$$

Remarks

The same formula holds for $x$ in any field, for example for complex numbers, when $x \neq 1$. For $x = 1$, the sum is $n$. A short proof: $(x - 1)(1 + x + \dots + x^{n-1}) = x^n - 1$, because the other terms cancel.

Explanations

L1 human

In a geometric progression, you get each term from the one before by multiplying by the same number $x$. For $x = 2$, the formula gives $1 + 2 + 4 + 8 = 2^4 - 1 = 15$. Multiply the sum by $x$ and subtract the sum: almost all terms cancel, and only $x^n - 1$ stays.