Sum of powers of \(q\)
An Introduction to Real Analysis
2025-10-03
Prove that for any number \(q \neq 1\), \[ \sum_{k = 0}^n q^k = \frac{q^{n + 1} - 1}{q - 1}. \] You can do this by induction, or by using identity (1.1.1).
Below are three proofs of this result: first by induction, second using identity (1.1.1), and lastly by a more intuitive geometric argument.
Proof (First). Since \[\begin{align*} \sum_{k = 0}^1 q^k &= q^1 + \sum_{k = 0}^0 q^k\\[0.75em] &= q + 1\\[0.75em] &= \frac{(q + 1)(q - 1)}{q - 1}\\[0.75em] &= \frac{q^{1 + 1} - 1}{q - 1}, \end{align*}\] the statement \(P(1)\) is true.1
Next, assuming that \(P(n)\) is true, \[\begin{align*} \sum_{k = 0}^{n + 1} q^k &= q^{n + 1} + \sum_{k = 0}^n q^k\\[0.75em] &= q^{n + 1} + \frac{q^{n + 1} - 1}{q - 1}\\[0.75em] &= \frac{q^{n + 1}(q - 1) + q^{n + 1} - 1}{q - 1}\\[0.75em] &= \frac{q^{n + 2} - q^{n + 1} + q^{n + 1} - 1}{q - 1}\\[0.75em] &= \frac{q^{(n + 1) + 1} - 1}{q - 1}, \end{align*}\] so \(P(n + 1)\) is true. \(\blacksquare\)
Proof (Second). Identity (1.1.1) is part of Proposition 1.1.3 in Katznelson and Katznelson (2024, 3). The proposition states that for any \(x, y \in \symbb{R}\) and any \(n \in \symbb{N}\), \[ x^n - y^n = (x - y) \sum_{k = 1}^n x^{n - k} y^{k - 1}. \] Relabeling \(x\) as \(q\) and taking \(y = 1\) and \(q \neq 1\) in this equation, \[\begin{align} q^n - 1 &= (q - 1)\sum_{k = 1}^n q^{n - k} \notag \\[0.75em] \implies \frac{q^n - 1}{q - 1} &= \sum_{k = 1}^n q^{n - k} \label{eq:1} \\[0.75em] &= \sum_{k = 0}^{n - 1} q^k. \label{eq:2} \end{align}\] In the last line the summation was re-ordered so the powers ascend.2 Adding \(q^n\) to \(\eqref{eq:2}\) and the left-hand side of \(\eqref{eq:1}\), \[\begin{align*} q^n + \sum_{k = 0}^{n - 1} q^k &= \frac{q^n - 1}{q - 1} + q^n\\[0.75em] \implies \sum_{k = 0}^n q^k &= \frac{(q^n - 1) + q^n(q - 1)}{q - 1}\\[0.75em] &= \frac{q^n - 1 + q^{n + 1} - q^n}{q - 1}\\[0.75em] &= \frac{q^{n + 1} - 1}{q - 1}, \end{align*}\] which completes the proof. \(\blacksquare\)
Proof (Third). Figure 1 gives the geometric inspiration for the third proof. Starting with \(\sum_{k = 0}^n q^k\), the figure suggests incrementing each power of \(q\) by \(1\) and considering the difference \(\sum_{k = 1}^{n + 1} q^k - \sum_{k = 0}^n q^k\). Then \[\begin{align} \sum_{k = 1}^{n + 1} q^k - \sum_{k = 0}^n q^k &= \overbrace{q^{n + 1} + \sum_{k = 1}^n q^k}^{= \sum_{k = 1}^{n + 1} q^k} - \overbrace{\left( \sum_{k = 1}^n q^k + 1 \right)}^{= \sum_{k = 0}^n q^k} \label{eq:3}\\[0.75em] &= q^{n + 1} - 1, \label{eq:4} \end{align}\] as suggested by the bottom row in Figure 1. Rewriting the left-hand side of \(\eqref{eq:3}\) to isolate \(\sum_{k = 0}^n q^k\), \[\begin{align} \sum_{k = 1}^{n + 1} q^k - \sum_{k = 0}^n q^k &= \sum_{k = 1}^{n + 1} qq^{k - 1} - \sum_{k = 0}^n q^k \notag\\[0.75em] &= q\sum_{k = 0}^n q^k - \sum_{k = 0}^n q^k \notag\\[0.75em] &= (q - 1)\sum_{k = 0}^n q^k. \label{eq:5} \end{align}\] Equating \(\eqref{eq:5}\) with \(\eqref{eq:4}\), \[\begin{align*} (q - 1)\sum_{k = 0}^n q^k &= q^{n + 1} - 1\\[0.75em] \implies \sum_{k = 0}^n q^k &= \frac{q^{n + 1} - 1}{q - 1}, \end{align*}\] which concludes the proof.3 \(\blacksquare\)
The geometric inspiration for the third proof, in Figure 1, tacitly assumes that \(q = 2\), or more generally that \(q \in \symbb{N}\setminus \{ 1 \}\). However, the proof makes no such assumption, so it is valid for \(q \in \symbb{R}\setminus \{ 1 \}\), just like the first two proofs. This is despite the fact that Figure 1 does not obviously apply when \(q = 4/3\) or \(q = -\uppi\), for example.