Sum of Geometric Progression/Corollary 1

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Corollary to Sum of Geometric Progression

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

Then:

$\displaystyle \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \left({r^n - 1}\right)} {r - 1}$


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.

(The Elements: Book $\text{IX}$: Proposition $35$)


Proof 1

By Multiplication of Numbers Distributes over Addition

$a + a r + a r^2 + \cdots + a r^{n - 1} = a \left({1 + r + r^2 + \cdots + r^{n - 1} }\right)$

The result follows from Sum of Geometric Progression.

$\blacksquare$


Proof 2

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

Hence:

\(\displaystyle \left({1 - r}\right) \sum_{0 \mathop \le j \mathop < n} a r^j\) \(=\) \(\displaystyle a - a r^n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{j \mathop = 0}^{n - 1} a r^j\) \(=\) \(\displaystyle \frac {a \left({1 - r^n}\right)} {1 - r}\)

The result follows.

$\blacksquare$


Historical Note

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


Sources