Sum of Sequence of Cubes/Proof by Induction

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

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $1$
1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (1) \ \text{(ii)}$
1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: Exercises: $1.2 \ \text {b)}$
1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $1 \ \text {(e)}$