# Sum of Geometric Sequence/Corollary 1/Proof 2

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## 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.*

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

## Proof

\(\ds \sum_{0 \mathop \le j \mathop < n} a r^j\) | \(=\) | \(\ds a + \sum_{1 \mathop \le j \mathop < n} a r^j\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a + r \sum_{1 \mathop \le j \mathop < n} a r^{j-1}\) | Distributive Property | |||||||||||

\(\ds \) | \(=\) | \(\ds a + r \sum_{0 \mathop \le j \mathop < n - 1} a r^j\) | 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.

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Example $3$. $(14)$