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:

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

$\blacksquare$