Quadratic Sequences

Gaps that grow

In a linear sequence the gap between one term and the next is always the same, so its points lie on a straight line. The sequence 3, 8, 15, 24 is different. It climbs by 5, then by 7, then by 9. Each step is bigger than the one before, so plotted on a graph the points would bend upward into a curve rather than lie on a line.

The gaps between the gaps

Write the gaps between neighboring terms. These are the first differences: 8 − 3 = 5, 15 − 8 = 7 and 24 − 15 = 9. Now find the gaps between the first differences: 7 − 5 = 2 and 9 − 7 = 2. These are the second differences.

The first differences change, but the second differences are all the same number, 2. When that happens, the sequence is quadratic: its nth term has an $n^2$n² in it.

Each term is the one before with a new block of dots beside it. The blocks hold 5, 7 and 9 dots, and each block is 2 bigger than the last.

Where the $n^2$n² comes from

Look at the square numbers 1, 4, 9, 16, 25. Their first differences are 3, 5, 7 and 9, and their second differences are all 2. So $n^2$n² on its own makes a second difference of 2. The sequence $2n^2$2n², which is 2, 8, 18, 32, makes a second difference of 4, and $3n^2$3n² makes 6. In general, when the coefficient of $n^2$n² is a, the second difference is 2a.

A part such as 2n, or a plain number, adds the same amount at every step. It changes every first difference by the same amount, so it leaves the second differences alone. That is why the second difference tells you about the $n^2$n² part and nothing else: the coefficient of $n^2$n² is half the second difference. Here the second difference is 2, so the coefficient is 2 ÷ 2 = 1, and the rule starts with $n^2$n².

$T_n$Tₙ is the nth term, ${\Delta }_1$Δ₁ is the row of first differences and ${\Delta }_2$Δ₂ is the row of second differences. Set a and b so that the terms run 3, 8, 15, 24. The second differences are 2 = 2a only when a = 1.

Take the $n^2$n² away

Now subtract $n^2$n² from each term and see what is left. At n = 1, 3 − 1 = 2. At n = 2, 8 − 4 = 4. At n = 3, 15 − 9 = 6. At n = 4, 24 − 16 = 8.

Each column takes $n^2$n² away from the term. What is left is 2, 4, 6, 8.

The part that is left

What is left, 2, 4, 6, 8, goes up by a steady 2, so it is a linear sequence. It starts at 2 and climbs by 2, so it is the 2 times table, 2n. Each term is $n^2$n² plus 2n, so the rule is $n^2+2n$n² + 2n.

Check the rule on a term you did not use to find it. At n = 5 the rule gives $5^2+2\times 5=25+10=35$5² + 2 × 5 = 25 + 10 = 35. The differences agree: the next first difference is 9 + 2 = 11, and 24 + 11 = 35.

What is left is not always a times table. If it is 3, 5, 7, 9, it climbs by 2 but starts at 3, so it is 2n + 1, found as in The nth Term.

Squares of tiles

In the next problem, the figures are square arrays of 4, 9 and 16 tiles. Their first differences are 5 and 7, and the second difference is 2, so the sequence is quadratic with $n^2$n² in its rule. Taking $n^2$n² away leaves 3, 5, 7, which is 2n + 1, so the rule is $n^2+2n+1$n² + 2n + 1.

The picture gives the same rule in a shorter form. Figure n is a square whose side is n + 1 tiles, so it holds ${\left(n+1\right)}^2$(n + 1)² tiles. Both forms give 676 for figure 25: $25^2+2\times 25+1=625+50+1=676$25² + 2 × 25 + 1 = 625 + 50 + 1 = 676, and $26^2=676$26² = 676.