C2.7; E2.7 - Square and cube sequences

Square-number patterns

A square made from tiles has side lengths $1,2,3,4,\ldots$. Its tile counts are $1,4,9,16,\ldots$. Why do the gaps grow by two each time, and how could you find the hundredth count without drawing a hundred squares?

The position number $n$ starts at $1$. A square with $n$ rows of $n$ tiles contains $n\times n=n^2$ tiles. The nth term gives the value at position $n$, rather than the amount added to the previous term.

[DIAGRAM: asset_name: Growing a square by an odd border; asset_slug: square_odd_border; caption: A three-by-three square gains seven tiles to become a four-by-four square.; description: Nine blue tiles form a three-by-three square. Four sage tiles extend its right edge and three more extend its top edge, creating a four-by-four square. The labels state old square nine tiles, new border seven tiles and total sixteen tiles.]

The next border has $5+4=9$ tiles. The differences between successive squares are the odd numbers $3,5,7,9,\ldots$: the square's two growing edges explain the pattern.

Compare terms at the same positions. The sequence $4,7,12,19,\ldots$ is not obtained by adding three repeatedly: its gaps are $3,5,7,\ldots$.

Change just the sign of the adjustment. The square-number gaps remain the same, although some early terms are now negative.

Cube-number patterns

A cube with side length $n$ small cubes has $n$ layers, each containing $n^2$ small cubes. Its total is $n\times n\times n=n^3$. The cube numbers are $1,8,27,64,125,\ldots$.

The cube-number gaps are $7,19,37,61,\ldots$, so adding the last gap forever would lose the pattern. An nth-term rule is more reliable than guessing a constant difference where there is none.