Square of Triangular Number equals Sum of Sequence of Cubes
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Contents
Theorem
- $\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.
Proof 1
| \(\displaystyle \sum_{i \mathop = 1}^n i^3\) | \(=\) | \(\displaystyle \frac {n^2 \paren {n + 1}^2} 4\) | Sum of Sequence of Cubes | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\frac {n \paren {n + 1} } 2}^2\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle {T_n}^2\) | Closed Form for Triangular Numbers |
$\blacksquare$
Proof 2
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
Basis for the Induction
$\map P 1$ is the case:
| \(\displaystyle \sum_{i \mathop = 1}^1 i^3\) | \(=\) | \(\displaystyle 1^3\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {\frac {1 \paren {1 + 1} } 2}^2\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle {T_1}^2\) | Closed Form for Triangular Numbers |
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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\displaystyle \sum_{i \mathop = 1}^k i^3 = {T_k}^2$
from which it is to be shown that:
- $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$
Induction Step
This is the induction step:
| \(\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3\) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle {T_k}^2 + \paren {k + 1}^3\) | Induction Hypothesis | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2}\) | Cube Number as Difference between Squares of Triangular Numbers | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle {T_{k + 1} }^2\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{> 0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
$\blacksquare$
