Sum of Sequence of Odd Squares/Formulation 2

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Theorem

$\displaystyle \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$


Proof

The proof proceeds by induction.

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

$\displaystyle \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_{i \mathop = 1}^1 \paren {2 i - 1}^2\) \(=\) \(\ds \paren {2 \times 1 - 1}^2\)
\(\ds \) \(=\) \(\ds 1^2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {4 \times 1 - 1}^3} 3\)

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown 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 the induction hypothesis:

$\displaystyle \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 = \frac {4 k^3 - k} 3$


from which it is to be shown that:

$\displaystyle \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2 = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3$


Induction Step

This is the induction step:

\(\ds \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2\) \(=\) \(\ds \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 + \paren {2 \paren {k + 1} - 1}^2\)
\(\ds \) \(=\) \(\ds \frac {4 k^3 - k} 3 + \paren {2 k + 1}^2\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {4 k^3 - k + 12 k^2 + 12 k + 3} 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1} \paren {4 k^2 + 8 k + 3} } 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1} \paren {4 \paren {k + 1}^2 - 1} } 3\)
\(\ds \) \(=\) \(\ds \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3\)

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


Therefore:

$\displaystyle \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$

$\blacksquare$


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