# Sums of k Powers

DevFeed: [Sums of k Powers](<https://devfeed.tech/articles/sums-of-k-powers-40216.md>)

Original publisher: [Read original article](<https://www.jeremykun.com/2011/07/01/sums-of-k-powers/>)

Published: 2011-07-01T12:52:29Z

Content type: article

Language: en

Sources: [Jeremy Kun](<https://devfeed.tech/sources/jeremy-kun.md>)

Topics: [polynomials](<https://devfeed.tech/topics/polynomials.md>), [context](<https://devfeed.tech/topics/context.md>)

Tags: [complex-numbers](<https://devfeed.tech/tags/complex-numbers.md>), [factoring](<https://devfeed.tech/tags/factoring.md>), [finite-fields](<https://devfeed.tech/tags/finite-fields.md>), [induction](<https://devfeed.tech/tags/induction.md>), [polynomial-ring](<https://devfeed.tech/tags/polynomial-ring.md>), [polynomials](<https://devfeed.tech/tags/polynomials.md>)

## AI overview

The article proves the geometric-series identity for sums of powers using base-k notation, polynomial factorization, and mathematical induction. It also explains extensions of the identity to fields such as the complex numbers and finite fields.

## Source excerpt

Problem: Prove that for all $ n,k \in \mathbb{N}, k > 1$, we have $$\sum \limits_{i=0}^{n} k^i = \frac{k^{n+1}-1}{k-1}$$ Solution: Representing the numbers in base $ k$, we have that each term of the sum is all 0's except for a 1 in the $ i$th place. Hence, the sum of all terms is the $ n$-digit number comprised of all 1's. Multiplying by $ k-1$ gives us the $ n$-digit number where every digit is $ k-1$.