More Quadratic Sequences

Section 1 of 2

When a is Bigger

Same method

The method never changes: $a = \tfrac{\text{2nd diff}}{2}$, then two equations from $T_1$ and $T_2$. $a$ can be a fraction or negative.

Find $T_n$ for $8, 14, 24, 38, \ldots$

Second difference $= 4$, so $a = 2$: $T_n = 2n^2 + bn + c$

$T_1$: $2 + b + c = 8 \Rightarrow b + c = 6$

$T_2$: $8 + 2b + c = 14 \Rightarrow 2b + c = 6$

Subtract: $b = 0$, $c = 6$

$T_n = 2n^2 + 6$

Find $T_n$ for $13, 15, 23, 37, \ldots$

Second difference $= 6$, so $a = 3$: $T_n = 3n^2 + bn + c$

$T_1$: $3 + b + c = 13 \Rightarrow b + c = 10$

$T_2$: $12 + 2b + c = 15 \Rightarrow 2b + c = 3$

Subtract: $b = -7$, then $c = 17$

$T_n = 3n^2 - 7n + 17$

Section 2 of 2

You try

Find $T_n$ for $6, 11, 25, 48, \ldots$

Second differences: $5, 14, 23 \to 9, 9$. So $a = 4.5$.

$a = \tfrac{9}{2} = 4.5$: $T_n = 4.5n^2 + bn + c$

$T_1$: $4.5 + b + c = 6 \Rightarrow b + c = 1.5$

$T_2$: $18 + 2b + c = 11 \Rightarrow 2b + c = -7$

Subtract: $b = -8.5$, then $c = 10$

$T_n = 4.5n^2 - 8.5n + 10$

You try

Find $T_n$ for $8, 10, 8, 2, -8, \ldots$

Second differences: $2, -2, -6, -10 \to -4, -4, -4$. So $a = -2$.

$a = \tfrac{-4}{2} = -2$: $T_n = -2n^2 + bn + c$

$T_1$: $-2 + b + c = 8 \Rightarrow b + c = 10$

$T_2$: $-8 + 2b + c = 10 \Rightarrow 2b + c = 18$

Subtract: $b = 8$, then $c = 2$

$T_n = -2n^2 + 8n + 2$

You try

Find $T_n$ for $68, 44, 28, 20, \ldots$

Second differences: $-24, -16, -8 \to 8, 8$. So $a = 4$.

$a = 4$: $T_n = 4n^2 + bn + c$

$T_1$: $4 + b + c = 68 \Rightarrow b + c = 64$

$T_2$: $16 + 2b + c = 44 \Rightarrow 2b + c = 28$

Subtract: $b = -36$, then $c = 100$

$T_n = 4n^2 - 36n + 100$

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