Sum of a Finite Geometric Series

The finite geometric series formula

$S_n = a_1 \cdot \frac{r^n - 1}{r - 1} \quad \text{(when } r \neq 1\text{)}$

Step-by-step: Find $1 + 2 + 4 + 8 + \cdots + 512$

Step 1 — Identify $a_1$, $r$, $n$: $a_1 = 1$, $r = 2$ (each term is double the previous), $n = 10$ terms (from $2^0 = 1$ to $2^9 = 512$).

Step 2 — Apply the formula:

$S = 1 \cdot \frac{2^{10} - 1}{2 - 1} = \frac{1024 - 1}{1} = 1023$

Step 3 — Check by adding the first few: $1 + 2 + 4 + 8 + 16 = 31$. Formula for 5 terms: $(2^5 - 1)/(2-1) = 31$ ✓.

Why the formula works

Multiply both sides of $S = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$ by $r$:

$rS = a_1 r + a_1 r^2 + \cdots + a_1 r^n$

Subtract: $S - rS = a_1 - a_1 r^n$, so $S(1 - r) = a_1(1 - r^n)$, giving $S = a_1 \frac{r^n - 1}{r - 1}$.

Neat observation

$1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1$. In binary, this is $111\ldots1$ ($n$ ones) = $2^n - 1$. Each power of 2 is one binary digit!

Try it in the visualization

Adjust $a_1$, $r$, $n$. The doubling bars show how each term grows. The running sum approaches the formula's prediction.