Sum of Geometric Progression/Proof 1

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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:

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

$\blacksquare$


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