Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Square_of_Triangular_Number_equals_Sum_of_Sequence_of_Cubes/Proof_2&oldid=535280"