Work out the formula of the nth term in the following quadratic sequence:

19, 15, 9, 1 ...

In a quadratic sequence we'll get a linear first difference and a constant second difference. Let's verify that.

n 1 2 3 4

f (n) 19 15 9 1

1st diff - 4 - 6 - 8

2nd diff 2 2

We see that we got a constant second difference. We could just extend that and work back up to get more values.

n 1 2 3 4 5 6 7

f (n) 19 15 9 1 - 9 - 21 - 35

1st diff - 4 - 6 - 8 - 10 - 12 - 14

2nd diff 2 2 2 2 2

That's just an aside; we're after the general formula. We have

f (1) = 19, f (2) = 15, f (3) = 9

In general we can assume

f (n) = an² + bn + c

We get three equations in three unknowns,

19 = a (1²) + b (1) + c = a+b+c

15 = a (2²) + b (2) + c = 4a + 2b + c

9 = a (3²) + b (3) + c = 9a + 3b + c

That's a 3x3 linear system; it's easy to solve directly. Subtracting pairs,

4 = - 3a - b

6 = - 5a - b

Subtracting those,

-2 = 2a

a = - 1

b = - 3a - 4 = - 1

c = 19-a-b = 21

Answer: f (n) = - n² - n + 21

Check:

f (1) = - 1 - 1 + 21 = 19, good

f (2) = - 4 - 2 + 21 = 15, good

f (3) = - 9 - 3 + 21 = 9, good

f (4) = - 16 - 4 + 21 = 1, good

Let's check our extended table, how about

f (7) = - 49 - 7 + 21 = - 35, good