# Sum of Geometric Sequence/Proof 4/Lemma

## Lemma

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

Then:

$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

## Proof

Proof by induction on $n$:

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

$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

### Basis for the Induction

$\map P 1$ is the case:

 $\ds \paren {1 - x} \sum_{i \mathop = 1}^{1 - 1} x^i$ $=$ $\ds 1 \paren {1 - x}$ $\ds$ $=$ $\ds 1 - x$ $\ds$ $=$ $\ds 1 - x^1$

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\ds \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i = 1 - x^k$

Then we need to show:

$\ds \paren {1 - x} \sum_{i \mathop = 0}^k x^i = 1 - x^{k + 1}$

### Induction Step

This is our induction step:

 $\ds \paren {1 - x} \sum_{i \mathop = 0}^k x^i$ $=$ $\ds \paren {1 - x} \paren {x^k + \sum_{i \mathop = 0}^{k - 1} x^i}$ $\ds$ $=$ $\ds x^k - x^{k + 1} + \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i$ $\ds$ $=$ $\ds x^k - x^{k + 1} + 1 - x^k$ from the induction hypothesis $\ds$ $=$ $\ds 1 - x^{k + 1}$ gathering terms

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

Therefore:

$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

$\blacksquare$