Sum of Sequence of Cubes/Proof by Induction

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Theorem

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


Proof

First, from Closed Form for Triangular Numbers:

$\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

So:

$\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$


Next we use induction on $n$ to show that:

$\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$


The proof proceeds by induction.

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

$\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$


Basis for the Induction

$\map P 1$ is the case:

$1^3 = \dfrac {1 \paren {1 + 1}^2} 4$


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


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:

$\ds \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$


from which it is to be shown that:

$\ds \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$


Induction Step

This is the induction step:

\(\ds \sum_{i \mathop = 1}^{k + 1} i^3\) \(=\) \(\ds \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3\)
\(\ds \) \(=\) \(\ds \frac {k^2 \paren {k + 1}^2} 4 + \paren {k + 1}^3\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {k^4 + 2 k^3 + k^2} 4 + \frac {4 k^3 + 12 k^2 + 12 k + 4} 4\)
\(\ds \) \(=\) \(\ds \frac {k^4 + 6 k^3 + 13 k^2 + 12 k + 4} 4\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2 \paren {k + 2}^2} 4\)

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


Therefore:

$\forall n \in \Z_{>0}: \ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$


Sources

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