Geometric series sum: the killer formula

The formula

For $q \neq 1$:

$1 + q + q^2 + \ldots + q^n = \sum_{k=0}^{n} q^k = \dfrac{1 - q^{n+1}}{1 - q}$

If $q = 1$: sum is $n + 1$ (every term = 1).

Standard form (start at 1)

$1 + q + q^2 + \ldots + q^n = \dfrac{1 - q^{n+1}}{1 - q}$

There are $n + 1$ terms.

For a geometric sequence

$u_n = u_0 \cdot q^n$:

$\sum_{k=0}^{n} u_k = u_0 \cdot \dfrac{1 - q^{n+1}}{1 - q}$

Infinite sum if $|q| < 1$

If $|q| < 1$, as $n \to +\infty$, $q^{n+1} \to 0$, so:

$\sum_{k=0}^{+\infty} q^k = \dfrac{1}{1 - q}$

Example: $1 + 1/2 + 1/4 + \ldots = 1/(1 - 1/2) = 2$.

Example

Compute $S = 1 + 2 + 4 + 8 + 16 + 32$.

$q = 2$, from $2^0$ to $2^5$, so $n = 5$.

$S = (1 - 2^6)/(1 - 2) = -63/-1 = 63$. ✓

Trick: term count

Sum from $q^p$ to $q^n$:

• Term count: $n - p + 1$
• Formula: $q^p \cdot \dfrac{1 - q^{n-p+1}}{1 - q}$

Common mistakes

1. $n$ vs $n+1$ in numerator: use $q^{n+1}$.
2. Forgetting the $q^0 = 1$ term: total $n+1$ terms.
3. Using the formula when $q = 1$: division by 0!

?À toi de jouer

What is 1 + 3 + 9 + 27 + 81?

A.121B.81C.243D.120

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