SUM OF CUBES

Abstract

A pictorial proof of the Sum of Cubes, that is
$$
\begin{array}{r}
4 \sum_{k=1}^n k^3=(3 n+1) \sum_{k=1}^n T_k+(3 n+2) \sum_{k=1}^{n-1} T_k \\
\left(T_k: \text { Triangular Number }\right),
\end{array}
$$
is provided. Since the square number is expressed as $k^2=T_k+T_{k-1}$, the cube number is expressed as $k^3=k T_k+k T_{k-1}$. For each cube number, we get two coefficients $k$ of this equation. And they can be placed in two tetrahedrons. And if we change the base of each tetrahedron four times, we get four tetrahedrons. By summing up the corresponding parts of these four tetrahedrons, the Sum of Cubes can be expressed as the formula shown above..

Received: December 19, 2022 
Accepted: January 5, 2023