# Sum of Geometric Progression/Proof 1

## 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:

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

## Proof

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:

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

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:

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

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

Therefore:

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

$\blacksquare$