Calculate the partial sum of any geometric series using Sₙ = a₁(1−rⁿ)/(1−r). Handles any common ratio including r = 1, negative r, and partial sums from term m to n.
Sₙ = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹
Write the sum twice — once multiplied by r:
Sₙ = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹
r·Sₙ = a₁r + a₁r² + … + a₁rⁿ⁻¹ + a₁rⁿ
Subtract the second row from the first. All middle terms cancel:
Sₙ(1−r) = a₁ − a₁rⁿ = a₁(1−rⁿ)
Divide both sides by (1−r) to get the final formula:
Sₙ = a₁(1−rⁿ) / (1−r), r ≠ 1
When r = 1 every term equals a₁, so Sn = n · a₁.
S∞ = a₁/(1−r), |r| < 1
As n → ∞, what happens to rⁿ?
If |r| < 1 (like r = 0.5), then rⁿ → 0, so the (1−rⁿ) factor approaches 1, and:
S∞ = a₁(1−0)/(1−r) = a₁/(1−r)
If |r| ≥ 1, the terms don't shrink — the series diverges (sum grows without bound).
This connection is why geometric series are so important: they're one of the few series types with a clean closed-form infinite sum.
One-on-one Algebra 2 tutoring walks through the derivation, the special cases, and the connection to infinite series — using your actual homework so the pattern sticks fast.