Quadratic Sequences — Mathslive.ie

Section 1 of 2

Spotting a Quadratic Sequence

The gap of the gap

If the differences are not constant but the differences of the differences (second differences) are, the sequence is quadratic: $T_n = an^2 + bn + c$.

Finding a, b, c

$a = \dfrac{\text{second difference}}{2}$. Then use $T_1$ and $T_2$ to make two equations and solve for $b$ and $c$.

Worked example — the square numbers

Find $T_n$ for $1, 4, 9, 16, 25, \ldots$

Second difference $= 2$, so $a = \tfrac{2}{2} = 1$: $T_n = n^2 + bn + c$

$T_1$: $1 + b + c = 1 \Rightarrow b + c = 0$

$T_2$: $4 + 2b + c = 4 \Rightarrow 2b + c = 0$

Subtract: $b = 0$, then $c = 0$

$T_n = n^2$

Worked example — a full quadratic

Find $T_n$ for $3, 7, 13, 21, \ldots$

Second difference $= 2$, so $a = 1$: $T_n = n^2 + bn + c$

$T_1$: $1 + b + c = 3 \Rightarrow b + c = 2$

$T_2$: $4 + 2b + c = 7 \Rightarrow 2b + c = 3$

Subtract: $b = 1$, then $c = 1$

$T_n = n^2 + n + 1$

You try

Find $T_n$ for $7, 16, 31, 52, \ldots$

Second differences: $9, 15, 21 \to 6, 6$. So $a = 3$.

$a = \tfrac{6}{2} = 3$: $T_n = 3n^2 + bn + c$

$T_1$: $3 + b + c = 7 \Rightarrow b + c = 4$

$T_2$: $12 + 2b + c = 16 \Rightarrow 2b + c = 4$

Subtract: $b = 0$, $c = 4$: $T_n = 3n^2 + 4$

$T_n = 3n^2 + 4$

That’s Class 8.

Second difference constant ⇒ quadratic; $a = \tfrac{\text{2nd diff}}{2}$, then solve for $b$ and $c$. Class 9: trickier quadratics.