Sum of Geometric Sequence/Corollary 1

Corollary to Sum of Geometric Sequence

Let $a, a r, a r^2, \ldots, a r^{n - 1}$ be a geometric sequence.

Then:

$\ds \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {r^n - 1} } {r - 1}$

In the words of Euclid:

If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it.

Proof 1

$a + a r + a r^2 + \cdots + a r^{n - 1} = a \paren {1 + r + r^2 + \cdots + r^{n - 1} }$

The result follows from Sum of Geometric Sequence.

$\blacksquare$

Proof 2

 $\ds \sum_{0 \mathop \le j \mathop < n} a r^j$ $=$ $\ds a + \sum_{1 \mathop \le j \mathop < n} a r^j$ $\ds$ $=$ $\ds a + r \sum_{1 \mathop \le j \mathop < n} a r^{j-1}$ Multiplication of Numbers Distributes over Addition $\ds$ $=$ $\ds a + r \sum_{0 \mathop \le j \mathop < n - 1} a r^j$ Exchange of Order of Summation $\ds$ $=$ $\ds a + r \sum_{0 \mathop \le j \mathop < n} a r^j - a r^n$

Hence:

 $\ds \paren {1 - r} \sum_{0 \mathop \le j \mathop < n} a r^j$ $=$ $\ds a - a r^n$ $\ds \leadsto \ \$ $\ds \sum_{j \mathop = 0}^{n - 1} a r^j$ $=$ $\ds \frac {a \paren {1 - r^n} } {1 - r}$

The result follows.

$\blacksquare$

Also presented as

This result is also seen presented as:

$\ds \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {1 - r^n} } {1 - r}$

which is usually more manageable when $r < 1$.

Historical Note

This proof is Proposition $35$ of Book $\text{IX}$ of Euclid's The Elements.