1730.32 – Squares and Cubes in a Pattern

First let us recognize that the strange numbers in the pattern, that is the numbers 1, 3, 6, 10 . . . are the sum of the first so many integers and that there is a formula for these, so-called triangular numbers, $T_n$:

$$\begin{aligned}T_1 &= 1 = 1 &= \dfrac{1 \cdot 2}{2} \\[1em]T_2 &= 3 = 1 + 2 &= \dfrac{2 \cdot 3}{2} \\[1em]T_3 &= 6 = 1 + 2 + 3 &= \dfrac{3 \cdot 4}{2} \\[1em]T_4 &= 10 = 1+2+3+4 &= \dfrac{4 \cdot 5}{2} \\[1em]T_5 &= 15 = 1+2+ \ldots +5 &= \dfrac{5 \cdot 6}{2} \\[1em]T_6 &= 21 = 1 +2 + \ldots +6 &= \dfrac{6 \cdot 7}{2}\end{aligned}$$

If we take $5^3$ as an example, we have

$$\begin{aligned}5^3 &= 15^2 - 10^2 \\&= T_5^2 - T_4^2 \\&= (T_5 + T_4)(T_5 - T_4) \\&= (15 + 10)(15 - 10) \\&= 25 \cdot5 = 5^3\end{aligned}$$

In general, we would like to have

$$n^3 = T_n^2 - T_n-1^2 = (T_n + T_n-1)(T_n - T_n-1)$$

We can recall that $T_n = 1 + 2 + \ldots + n = n (n + 1)/2.$

Thus

$$\begin{aligned}T_n + T_{n-1} &= \dfrac{n (n+1)}{2} + \dfrac{(n - 1) n}{2} \\&= \dfrac{n^2 + n + (n^2 - n)}{2} \\&= \dfrac{2 n^2}{2} = n^2\end{aligned}$$

Similarly,

$$\begin{aligned}T_n - T_{n-1} &= \dfrac{n (n+1)}{2} - \dfrac{(n - 1) n}{2} \\&= \dfrac{n^2 + n - (n^2 - n)}{2} \\&= \dfrac{2 n}{2} = n\end{aligned}$$

So it works!

$$\begin{aligned}T_n^2 - T_{n-1}^2 &= (T_n - T_{n-1})(T_n + T_{n-1}) \\&= n \cdot n^2 \\&= n^3.\end{aligned}$$

Voila!