Finding Sum of Geometric Series

Algebra 2 · Axiom Academy

LESSON Finding the Sum of a Geometric Series Derive the closed form Sₙ = a₁(1 − rⁿ)/(1 − r) with one clever move: multiply the whole sum by r and subtract. 1. The Problem: a Sum That Grows Fast The sum of the first n terms of a geometric series is n terms, exponents running 0 to n − 1 Every term is just the one before it multiplied by the common ratio r: 2. The Key Trick: Multiply by r, Then Subtract Write the sum Sₙ. Now write rSₙ — the exact same terms, but each bumped up one power of r, so the whole line is shifted one place to the right : Line them up. Every interior term of Sₙ sits directly above an identical term of rSₙ, so subtracting cancels them all — only the first term of Sₙ and the last term of rSₙ are left standing: Factor Sₙ out of the left side and a₁ out of the right: Then divide both sides by (1 − r) — valid whenever r ≠ 1: the sum of the first n terms, in closed form Multiplying top and bottom by −1 gives the equivalent form often used when r > 1: The division needed r ≠ 1. In the leftover case r = 1 every term equals a₁, so the sum is simply Use the formula on 2 + 6 + 18 + 54 + 162. Reading off the pieces: Now check the shortcut against plain addition: You derived the geometric-series sum formula from scratch — the entire proof is one multiplication and one subtraction. Scroll up to revisit any step.

This is the written version of the interactive lesson above. See the full Algebra 2 course.