Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2
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Theorem
- $\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.
Proof
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sum_{i \mathop = 1}^1 i^3\) | \(=\) | \(\ds 1^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {1 \paren {1 + 1} } 2}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {T_1}^2\) | 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:
- $\ds \sum_{i \mathop = 1}^k i^3 = {T_k}^2$
from which it is to be shown that:
- $\ds \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$
Induction Step
This is the induction step:
\(\ds \sum_{i \mathop = 1}^{k + 1} i^3\) | \(=\) | \(\ds \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {T_k}^2 + \paren {k + 1}^3\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2}\) | Cube Number as Difference between Squares of Triangular Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds {T_{k + 1} }^2\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{> 0}: \ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$