ThmDex – An index of mathematical definitions, results, and conjectures.

Real arithmetic expression for sum of a finite real geometric sequence

Formulation 0

Let $q \in [0, 1)$ be an D4767: Unsigned real number.
Let $N \in 1, 2, 3 \ldots$ be a D5094: Positive integer.

Then \begin{equation} \sum_{n = 0}^N q^n = \frac{1 - q^{N + 1}}{1 - q} = \frac{q^{N + 1} - 1}{q - 1} \end{equation}

Formulation 1

Let $q \in [0, 1)$ be an D4767: Unsigned real number.
Let $N \in 1, 2, 3 \ldots$ be a D5094: Positive integer.

Then \begin{equation} q^0 + q^1 + q^2 + \cdots + q^N = \frac{1 - q^{N + 1}}{1 - q} = \frac{q^{N + 1} - 1}{q - 1} \end{equation}

Proofs

Proof 0

Let $q \in [0, 1)$ be an D4767: Unsigned real number.
Let $N \in 1, 2, 3 \ldots$ be a D5094: Positive integer.

We can write \begin{equation} \begin{split} q^{N + 1} - 1 = q^{N + 1} - q^0 & = \left( \sum_{n = 1}^{N + 1} q^n \right) - \left( \sum_{n = 0}^N q^n \right) \\ & = q \left( \sum_{n = 0}^N q^n \right) - \left( \sum_{n = 0}^N q^n \right) \\ & = (q - 1) \left( \sum_{n = 0}^N q^n \right) \end{split} \end{equation} Dividing each side by the nonzero quantity $q - 1$, we then get \begin{equation} \left( \sum_{n = 0}^N q^n \right) = \frac{q^{N + 1} - 1}{q - 1} = \frac{(-1) (1 - q^{N + 1})}{(-1) (1 - q)} = \frac{1 - q^{N + 1}}{1 - q} \end{equation} $\square$