# Sum of Geometric Progression

## Theorem

Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.

Let $n \in \N_{>0}$.

Then:

$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$

### Corollary 1

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

Then:

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

### Corollary 2

$\displaystyle \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\left({n - 1}\right) x^{n + 1} - n x^n + x} {\left({x - 1} \right)^2}$

## Proof 1

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$

### Basis for the Induction

$\map P 1$ is the case:

$x^0 = \dfrac {x^1 - 1} {x - 1} = 1$

so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\displaystyle \sum_{j \mathop = 0}^{k - 1} x^j = \frac {x^k - 1} {x - 1}$

Then we need to show:

$\displaystyle \sum_{j \mathop = 0}^k x^j = \frac {x^{k + 1} - 1} {x - 1}$

### Induction Step

This is our induction step:

 $\displaystyle \sum_{j \mathop = 0}^k x^j$ $=$ $\displaystyle \sum_{j \mathop = 0}^{k - 1} x^j + x^k$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x^k - 1} {x - 1} + x^k$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x^k - 1 + \paren {x - 1} x^k} {x - 1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x^k - 1 + x^{k + 1} - x^k} {x - 1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x^{k + 1} - 1} {x - 1}$ $\quad$ $\quad$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$

$\blacksquare$

## Proof 2

Let $\displaystyle S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.

Then:

 $\displaystyle \left({x - 1}\right) S_n$ $=$ $\displaystyle x S_n - S_n$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle x^n + \sum_{j \mathop = 1}^{n - 1} x^j - \left({x^0 + \sum_{j \mathop = 1}^{n - 1} x^j}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle x^n - x^0$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle x^n - 1$ $\quad$ $\quad$

The result follows.

$\blacksquare$

## Proof 3

$\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} }\right) = \left({a - b}\right) \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$

Set $a = x$ and $b = 1$:

$\displaystyle x^n - 1 = \left({x - 1}\right) \left({x^{n - 1} + x^{n - 2} + \cdots + x + 1}\right) = \left({x - 1}\right) \sum_{j \mathop = 0}^{n - 1} x^j$

from which the result follows directly.

$\blacksquare$

## Proof 4

### Lemma

Let $n \in \N_{>0}$.

Then:

$\displaystyle \left({1 - x}\right) \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

$\Box$

Then by the lemma:

 $\displaystyle \left({1 - x}\right) \sum_{i \mathop = 0}^{n - 1} x^i$ $=$ $\displaystyle 1 - x^n$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}$ $=$ $\displaystyle \frac {x^n - 1} {x - 1}$ $\quad$ $\quad$

$\blacksquare$

## Also presented as

Note that when $x < 1$ the result is usually given as:

$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {1 - x^n} {1 - x}$

Some sources give it as:

$\displaystyle \sum_{j \mathop = 0}^n x^j = \frac {1 - x^{n + 1} } {1 - x}$

and likewise its corollary:

$\displaystyle \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$

## Examples

### $\dfrac 1 7$ from $1$ to $n$

$\displaystyle \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \left({1 - \frac 1 {7^{n + 1} } }\right)$

### Common Ratio $1$

Consider the Sum of Geometric Progression defined on the standard number fields for all $x \ne 1$.

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

When $x = 1$, the formula reduces to:

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

### Index to $-1$

Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.

Then the formula for Sum of Geometric Progression:

$\displaystyle \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$

still holds when $n = -1$:

$\displaystyle \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$

### Index to $-2$

Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.

Then the formula for Sum of Geometric Progression:

$\displaystyle \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$

breaks down when $n = -2$:

$\displaystyle \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$