Sum of Sequence of Odd Cubes
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Theorem
- $\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$
Proof
Proof by induction:
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
- $\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sum_{j \mathop = 1}^1 \paren {2 j - 1}^3\) | \(=\) | \(\ds 1^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1^2 \paren {2 \times 1^2 - 1}\) |
and $\map P 1$ is seen to hold.
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:
- $\ds \sum_{j \mathop = 1}^k \paren {2 j - 1}^3 = k^2 \paren {2 k^2 − 1}$
Then we need to show:
- $\ds \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}^3 = \paren {k + 1}^2 \paren {2 \paren {k + 1}^2 − 1}$
Induction Step
This is our induction step:
\(\ds \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}^3\) | \(=\) | \(\ds \sum_{j \mathop = 1}^k \paren {2 j - 1}^3 + \paren {2 k + 1}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^2 \paren {2 k^2 − 1} + \paren {2 k + 1}^3\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 k^4 + 8 k^3 + 11 k^2 + 6 k + 1\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^2 \paren {2 k^2 + 4 k + 1}\) | extracting $\paren {k + 1}^2$ as a factor | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^2 \paren {2 \paren {k + 1}^2 - 1}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 1}: \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$
$\blacksquare$