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
